Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been struggling with general topology and now, algebraic topology is simply murder. Some people seem to get on alright, but I am not one of them unfortunately.

Please, an answer I need is something very elaborate and extensive, possibly with easy-to-understand examples. Rewriting the definitions with concrete mathematical language and symbols won't help(those are readily accessible in my lecture notes). I want an explanation of what is happening and which part of the solid definitions are telling me so.

One thing I would like to make clear is, I know the definitions I don't understand them; I can re-iterate them upon request, consult my lecture notes. Issue is I am writing down something I don't know what it means. It's like writing ancient Greek. I can remember the shapes of each character, their order and write some sentence down. But that's it. Can I explain it to someone with my own words? Break it down? Absolutely not. That is why I am here to ask someone who fully understands these ideas to do exactly that for me: break it down. Go in slow motion. Show me the moves and codes behind it.

Here's the definition I have for homotopy

A homotopy between maps $f,g:X \rightarrow Y$ is a map $h: X \times I \rightarrow Y$ such that

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ where $x \in X$ and $I=[0,1]$. We say maps $f,g$ are homotopic.

It adds that,

A homotopy deforms the map $f$ continuously to $g$.

So, I am given two sets $X,Y$ whatever they are and two maps $f,g$ that takes an element of $X$ to an element of $Y$. I don't know if $f,g$ are bijective, only injective or surjective or whatever. No information on that. Just maps.

And this "homotopy" is a ... "map between maps"? And even if so, what exactly is it doing? All the explanation seems to be done with this single line

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$

but no, I don't get what's happening. So $X$ has a bunch of elements $x$, and the product space $X \times I$ gives me elements of the form $\{x,t\} \in X \times I$. Okay. But this map $h$ qualifies as a homotopy as long as the above holds? Then why not just always define $h$ as $h(x,t)=f(x)$ for any $t \neq 1$ and $h(x,t)=g(x)$ for $t=1$, just as how we might define a piecewise function? Then I can define this "homotopy" on any maps.

Sure, it adds "deforms $f$ CONTINUOUSLY to $g$" but how is that stated in the definition itself? I see it nowhere.

Here is an example in my notes which didn't help me understand the definition,

Take $X=\{x\}$ the space with a single element $x$. Then a map $f:X \rightarrow Y$ is the same as and element $f(x) \in Y$. A homotopy $h: f \simeq g: X \rightarrow Y $ is the same as a path $h: I \rightarrow Y$ with initial point $h(0)=f(x)$ and terminal point $h(1)=g(x) \in Y$. A homotopy $h: f \simeq f: X \rightarrow Y$ is the same as a closed path $h: I \rightarrow Y$.

Well, first off what is a "path"? Intuition also doesn't make sense because when mapping one element to another, how can there be different "paths"? $x$ goes to $y$. Done. It's not like going from England to Singapore via either Amsterdam or Frankfurt (thus different paths) is it? Unless it's a map $X \rightarrow Z \rightarrow Y$ and telling me that $x \in X$ goes to $z_1 \in Z$ then to $y \in Y$ or $z_2 \in Z$ and then to $y \in Y$, that might be different paths from $x$ to $y$. But here, it is talking about $X$ and $Y$ only.

And why is this ignoring $x$? It says $h: I \rightarrow Y$? The problem I also have is, this is labelled "example" but it's not specific at all. "$h(0)=f(x)$ and $h(0)=g(x)$" over. So? What is this $h$ how has it been defined? As a map, as a function of some sort? Maybe there are multiple homotopy thinkable, but then what are one or two of them?

It's like saying $f(1)=1$ and $f(2)=4$. Done. Well, to a newbie, maybe secondary school kids, it might be nice to give them an example be it linear $f(x)=ax+b$ or $f(x)=x^2$. To the eyes of the experienced, it might make perfect sense, a specific homotopy might pop out in their minds like popcorns but not in mine.

It's an utter nightmare. I know this is "abstract" math but can not some more specific-ness be put into it? Pictures and diagrams perhaps?

This is only the tip of the massive massive confusion and dumb-foundedness I am experiencing in this area of study. Maybe once it "clicks" it all goes down like an avalanche but so far it's nothing but counter-intuitive.

Can someone please make this possible for me to digest? Suggestions to good websites with examples and diagrams and extensive explanations are also welcome. Thank you

share|cite|improve this question

closed as too broad by Pedro Tamaroff Feb 12 at 15:23

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

4  
Roughly speaking: two paths from $a$ to $b$ are homotopically equivalent if you can colour in the gap between them. In $\mathbb R^n$ you can always do this. But in $\mathbb R^2 \setminus (0,0)$, there might be a hole in the space between your two paths. Wikipedia has a helpful graphic. – Mathmo123 Feb 10 at 11:42
3  
Those tempted to answer in a 600-character comment should consider rereading the opening paragraphs of the post. – user1717828 Feb 10 at 16:49
2  
For that definition to be correct, "map" needs to mean "continuous function". ​ ​ – Ricky Demer Feb 10 at 18:44
2  
The title becomes much stranger if you switch the quotations from homotopy to "me". – Asaf Karagila Feb 10 at 21:42
4  
Duchamp's Nude Descending a Staircase is a famous visualization of (the range of) a homotopy between a nude at the top of some stairs, and a nude at the bottom of some stairs. – enthdegree Feb 11 at 18:57

10 Answers 10

up vote 16 down vote accepted

I sympathize with your situation - I also struggled a lot with algebraic topology. I'll try to help you answering specific questions.

So, I am given two sets $X,Y$ whatever they are and two maps $f,g$ that takes an element of $X$ to an element of $Y$. I don't know if $f,g$ are bijective, only injective or surjective or whatever. No information on that. Just maps.

Since we're in algebraic topology, you can safely assume that $X$ and $Y$ are topological spaces, and $f$ and $g$ are continuous maps. We really don't have information about injectiveness/surjectiveness of the maps involved. We can define "homotopy" in another contexts, such as linear algebra, group theory, etc, too, but I'll get to that later.

And this "homotopy" is a ... "map between maps"? And even if so, what $exactly$ is it doing? All the explanation seems to be done with this single line $$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ but no, I don't get what's happening. So $X$ has a bunch of elements $x$, and the product space $X \times I$ gives me elements of the form $\{x,t\} \in X \times I$. Okay. But this map $h$ qualifies as a homotopy as long as the above holds? Then why not just always define $h$ as $h(x,t)=f(x)$ for any $t \neq 1$ and $h(x,t)=g(x)$ for $t=1$, just as how we might define a piecewise function? Then I can define this "homotopy" on $any$ maps.

If you define $h$ as you said above, most likely it will not be continuous at the pairs $(x,1)$, we have a "jump" in $t$ near that. I'd guess that the lecturer there probably said something along the lines of "in this course, unless otherwise stated, all maps will be continuous". So it is not true that there is a homotopy between any maps.

I will give three examples of homotopies and then try to give the general idea:

  • Let $f,g\colon X \to \Bbb S^2$ be continuous maps such that $f(x) \neq -g(x)$ for all $x \in \Bbb S^2$, where $X$ is any topological space. Define $h\colon X \times I \to \Bbb S^2$ by $$h(x,t) = \frac{(1-t)f(x)+tg(x)}{\|(1-t)f(x)+tg(x)\|}.$$We are not dividing by zero because of the $f(x) \neq -g(x)$ part (the problem would be for $t = 1/2$). Check that this is a homotopy to get the feeling for the computations (i.e., $h(x,0) = f(x)$, $h(x,1) = g(x)$, and $h$ is continuous).

  • Let $f\colon \Bbb R^n \to \Bbb R^n$ be a continuous map. Then $h\colon \Bbb R^n \times I \to \Bbb R^n$ given by $h(x,t) = tx + (1-t)f(x)$ is a homotopy between $f$ and the identity function (check that, same as above).

  • Let $f\colon \Bbb R^n \to \Bbb R^k$ be a continuous maps. Then $h\colon \Bbb R^n \times I \to \Bbb R^k$ given by $h(x,t) = tf(x)$ is a homotopy between $f$ and the zero map.

A homotopy is a map between maps in the following sense: fix $f,g\colon X \to Y$ continous maps, and $h\colon X \times I \to Y$ the homotopy. We can see $h$ as the map that takes $t \in I$ to $h_t \in {\cal C}(X,Y)$, where $h_t$ is defined by $h_t(x) = h(x,t)$. Going up a bit ahead, we'll say that a path in a topological space is a map defined on $I$ taking values in the space. Then the above interpretation sees $h$ as a path in ${\cal C}(X,Y)$, connecting $f$ and $g$, since $h_0 = f$ and $h_1 = g$, as functions.

A "dual" interpretation is seen by fixing $x$, and seeing $h$ as the map that takes $x \in X$ to the map $h_x \in {\cal C}(I,Y)$, given by $h_x(t) = h(x,t)$. Then for every fixed $x$, $h_x$ is a path in $Y$, connecting $f(x)$ and $g(x)$.

Here are some notes from where I took the course (in portuguese, but I want a specific image there) - in the first page, there is a square with $f$ in the bottom, $g$ in the top, the left side labeled $I$, etc. The maps $h_t$ are constant in horizontal lines, and the $h_x$ are constant in vertical lines, try to think a bit about that.

Sure, it adds "deforms $f$ CONTINUOUSLY to $g$" but how is that stated in the definition itself? I see it nowhere.

Saying that the deformation is continuous is just jargon for saying that $h\colon X \times I \to Y$ is continuous (we consider the product topology in $X \times I$ here). If the lecturer made the observation of all maps being continuous otherwise stated, as I speculated above, then I think this issue here is solved.

Well, first off what is a "path"? Intuition also doesn't make sense because when mapping one element to another, how can there be different "paths"? $x$ goes to $y$. Done. It's not like going from England to Singapore via either Amsterdam or Frankfurt (thus different paths) is it? Unless it's a map $X \rightarrow Z \rightarrow Y$ and telling me that $x \in X$ goes to $z_1 \in Z$ then to $y \in Y$ or $z_2 \in Z$ and then to $y \in Y$, that might be different paths from $x$ to $y$. But here, it is talking about $X$ and $Y$ only.

As I rushed up before: a path in a topological space $X$, connecting $x$ and $y$ in $X$, is a continuous maps $\gamma\colon I \to X$ such that $\gamma(0) = x$ and $\gamma(1)=y$. It is possible that given $x,y \in X$, there is more than one path connecting them, or that there is no path connecting them, at all. Given continuous maps $f,g$ between topological spaces, there could be more than one homotopy between them, or no homotopy at all.

In general topology, you should've met the concept of a "connected space". We'll say that a topological space is "path-connected" if any two points can be connected by (at least) one path. Every path-connected space is connected, the converse is false. The idea of algebraic topology, at a first moment, is to distinguish topological spaces (modulo homeomorphism) by topological invariants - and the focus is on connectedness/path-connectedness, and one of the most useful instruments to do that depends on homotopies.

Concrete examples of paths:

  • given two points in $\Bbb R^2$, we can join them by a straight line, or by a half-circumference, etc.

  • if $X = [0,1] \cup [2,3]$, and if $x \in [0,1]$, $y \in [2,3]$, there is no continuous path between $x$ and $y$ - by the intermediate value theorem, such a path $\gamma$ would have to assume the value, say, $1.5$, which is not in $X$ (the problem is that the path passes outside $X$, and we do not want that).

  • In $\Bbb S^1 = \{e^{it} \mid t \in \Bbb R\}$, if we take $x = e^{it_0}$ and $y = e^{it_1}$, then $\gamma\colon I \to \Bbb S^1$ given by $\gamma(t) = e^{i((1-t)t_0+tt_1)}$ verifies $\gamma(0) = x$ and $\gamma(1) = y$. This is one possible path between them, going in one direction in the circle - you could go the other way.

Take $X={x}$ the space with a single element $x$. Then a map $f:X→Y$ is the same as and element $f(x)∈Y$. A homotopy $h:f≃g:X→Y$ is the same as a path $h:I→Y$ with initial point $h(0)=f(x)$ and terminal point $h(1)=g(x)∈Y$. A homotopy $h:f≃f:X→Y$ is the same as a closed path $h:I→Y$.

This is the interpretation with $h_x$ that I gave above, but since there only is one $x \in X$, they abused notation and wrote $h\colon I \to Y$, where more precisely we'd write $h_x\colon I \to Y$.

I hope this helps a bit, even. And you can ask away stuff in the comments.


Extra: homotopies in another contexts;

  • In linear algebra: given vector spaces $V,W$, and linear maps $T,S\colon V \to W$, a homotopy between $T$ and $S$ is a map $H\colon V \times I \to W$ such that $H(x,0) = T(x)$, $H(x,1) = S(x)$, and for all $t \in I$, the maps $H_t(x) = H(x,t)$ is linear;

  • In group theory: given groups $G,H$, and group homomorphisms $\varphi, \psi\colon G \to H$, a homotopy between $\varphi$ and $\psi$ is a map $h\colon G \times I \to H$ such that $h(x,0) = \varphi(x)$, $h(x,1) = \psi(x)$, and for all $t \in I$, $h_t(x) = h(x,t)$ is a group homomorphism;

And so on. If you know a bit of category theory, you see the pattern: we want a $1$-parameter family of morphisms joining the initial ones. For topological spaces, a $1$-parameter family of continuous maps; for vector spaces, a $1$-parameter family of linear maps, etc.

share|cite|improve this answer
1  
Wow! Explanation of the "duality" and giving some taste of generalization on the level of category theory gave me new unexpected understanding for the subject. Thank you. – Mihail Feb 10 at 20:06
1  
"$h(x,t)=tx+(1−t)f(x)h(x,t)=tx+(1−t)f(x)$ is a homotopy between $f$ and the identity" would be a clearer phrase, IMHO, with the addition of the word "function" after the word "identity". It took me a moment to understand that $h$ did not become everywhere $1$ as $t$ goes to $1$. – Todd Wilcox Feb 10 at 21:00
1  
@Todd ok, fair. Fixed :) – Ivo Terek Feb 10 at 21:41

Imagine a malleable object like a rubber band that can occupy different shapes/configurations in space.

Let us call the abstract reference rubber band $X$ (independent of any specific configuration it might take on), and call the space it moves in $Y$.

To define a particular configuration of the rubber band, one would need to specify where each point in the reference rubber band exists in space. That is, each possible configuration of the rubber band in space corresponds to a map from $X$ to $Y$.

Maps $f,g:X \rightarrow Y$ then correspond to two different configurations of the rubber band in space.

Maps $h:X \times I \rightarrow Y$ correspond to a time-sequence of configurations that the rubber band takes on. For every point $x$ in the reference rubber band and for every time $t$ in $I$, the location of that part of the rubber band in space at that time is specified by $h(x,t)$. Visualize $h$ as a movie of the rubber band moving and deforming in space over time.

Now, nothing written so far prevents crazy stuff from happening like the map $f$ taking the rubber band and mapping it to some weird disconnected dust or something like that. Similarly, nothing yet written prevents the rubber band in the movie $h$ to spontaneously teleport from one place to another at every frame, or things like that. The continuity assumptions are minimal assumptions there to prevent such crazy things.

Note that in particular, the continuity assumption still allows the rubber band to pass through itself, or to squish down from a 3d volumetric band to a 2D ring, so long as these processes are done in a continuous manner. If one desires to prevent these cases as well, then more stringent requirements can be used (e.g., isotopy, where the framework is the same but the maps are also required to be injective)

A natural question is, given two continuous configurations $f$ and $g$ of the rubber band, is it possible to continuously move/bend/twist/squish one into the other via a continuous movie $h$? This is equivalent to asking whether there exists a homotopy $h$ between configurations $f$ and $g$.

Hope this helps with the intuition.

share|cite|improve this answer
15  
In my view there is an overemphasis on rigor and precision when introducing students to topics such as this. This type of explanation should be in textbooks, followed by a discussion of the formal definition and why it properly captures the intuition. At least in my (somewhat limited) experience, only after one has grasped the intuitive idea are the rigorous definitions motivated or useful. – Will Feb 11 at 0:38
    
One pretty well-known introduction to algebraic topology is by Hatcher (math.cornell.edu/~hatcher/AT/AT.pdf ). His Chapter 0 "Some underlying geometric notions" gives the geometric intuition for a lot of the later concepts. – Chan-Ho Suh Feb 11 at 15:09
1  
@Will: I disagree. The point of definitions is not necessarily to capture intuition. Look at CW complexes, for example: Their construction is somewhat intuitive, but the reason they're defined that way is to avoid pathologies (enough for, say, the Whitehead theorem to hold) but still be general enough to be reasonable easy to work in (as opposed to, say, simplicial complexes). It's important to note the motivation and reason for the definition, but those aren't the same as intuition. (And often disagree; the rubber band intuition is closer to isotopy than homotopy.) – anomaly Feb 11 at 18:00
1  
@Will I agree in principle, except I would say the intuitive explanations (while extremely important) are usually best left to the instructor, either to explicate during lecture or assign as an exercise depending on the sophistication of the class. – Todd Wilcox Feb 11 at 19:33
    
@ToddWilcox, I agree in principle ;) In my experience, however, effective instructors who can deliver an intuitive explanation are a rare luxury. In fact, having an instructor at all is a luxury. Solid, rigorous reference texts are of course essential—but I think there is also a need for texts that use explanations like the above. Too often the instructor simply regurgitates the definitions in the standard textbooks, as evidenced by OP's frustration. – Will Feb 11 at 21:21

I'm surprised noone presented a drawing. A homotopy (between paths) can be naively represented as follows.

Suppose we have a path $c_0$ in space. Like this:

enter image description here

We now move it around, indexing its movement on the subscript. For instance,

enter image description here

Making this a continuous process, we get a function $G:I \rightarrow \mathcal{F}(I, Y); \quad s \mapsto c_s$, which induces in a clear manner a function $F: I \times I \rightarrow Y$ by $(s,t) \mapsto c_s(t)$.

A homotopy between maps is simply not restricting ourselves to paths. We consider the maps we want to deform to be maps $f: X \rightarrow Y$, and we get a function $G: I \rightarrow \mathcal{F}(X, Y)$ which induces $F:X \times I \rightarrow Y$ in the same way.

share|cite|improve this answer
7  
Sidenote: In my personal opinion, "trivial" examples are commonly more confusing than enlightening ones, mainly for students. If you want to show an example of a structure, start by showing a interesting and simple one, not a trivial one, since it may not illustrate the structure. I think this happened to you. It is similar to presenting as first example of a path a constant path, instead of a straight line. – Aloizio Macedo Feb 10 at 15:25
    
This would be clear if you add a statement before the final paragraph: which thing is the homotopy between paths? You don't say. I'm thinking maybe the map C0 -> C1 or maybe the parameterized map C0 -> Ci where i is a continuous variable. – Alan Baljeu Feb 11 at 18:34

There's a pedagogical quirk that makes homotopies extremely easy to explain, and that's that they can (should?) be thought of as occurring over time, and there's this wonderful invention called a video that very naturally expresses how shapes can change over time.

Try Regular homotopies in the plane... it's some sort of cult classic but I recommend it with no irony for your question.

share|cite|improve this answer
1  
+1 The linked video is about regular homotopies, which have the added condtion of having continuous derivatives, so the result is not really about homotopy itself. But if one ignores the talk about tangent vectors, it still gives a good intuitive feel for general homotopies. – Paul Sinclair Feb 11 at 2:20
1  
You linked to a German dub. (There's a link to the English version in the description.) – Akiva Weinberger Feb 11 at 23:56
    
@AkivaWeinberger fixed. – Thomas Feb 12 at 11:33

A lot of good answers were already written—I just wanted to add something that I don't see and that I found really, really useful when first studying this: why did we introduce homotopy in the first place? As often happens in mathemathics once you understand what's the idea behind something, the mathemathical formalism suddenly turns easy and meaningful.

All this stuff was introduced by Poincaré in a single article to study surfaces. The main issue here (explained in layman's terms) is that you want to understand if two surfaces are in fact the same from a topological point of view. Let us limit ourselves for the moment to surfaces to get to the heart of the matter.

Poincaré studied the problem for a while and understood that two surfaces were topologically equivalent if there were two continuous functions from one surface to the other, respectively (i.e. homeomorfic).

This was the right condition, but let's say you can't determine if there are two continuous functions between two surfaces—is that enough to say that they are not equal? Indeed, maybe it's just that you can't find the right functions… so what did Poincaré think? Well, let's just put a curve on the surface, i.e. instead of considering the whole surface, let us just consider curves on the surface.

What's going on is that if there are two continuous functions between the two surfaces there must be two continuous functions that deform a curve on one surface into the curve on the other. And so here's the catch: this means that if I consider all the continuous deformations of one curve on a surface (the homotopy class of the curve) this must correspond to all the continuous deformations of the curve on the other surface (i.e. a homotopy class on the other surface).

So Poincaré calculated the homotopy group (i.e. the group formed by the homotopy classes of the curves) of one surface and the homotopy group of the other, and saw that they were not the same—he then deduced that the two surfaces could not be deformed in a continuous way one into the other.

All the definitions you have are just the mathematical formalization of this picture. And if you have this picture you can also generalize the procedure for volumes, by considering the deformations of surfaces (second homotopy group), and further with deformation of volumes for hypervolumes, etc.

share|cite|improve this answer

I guess this question makes arise another important question: what does it mean to understand something? I leave someone else the task to answer this.

Here some idea that I hope may help you in getting more confident with basic concepts of homotopy theory.

I believe we can start from the most basic concept which is the one of path. By definition a path, in a topological space $X$, is nothing but a continuous function from the unit interval (i.e. mr. $I=[0,1]$) into a space $X$: that is an object of the form $\alpha \colon I \to X$ that is continuous.

The definition gives more or less everything you need to know in order to work with paths. So a path is something that associates to each real number between $0$ and $1$ a point in the space $X$ (in a continuous way). There is nothing else to understand other than this.

Of course there are some ideas behind this definition. For instance you can think, dynamically, a path $\alpha$ as being a family of points you have to move on in order to reach $\alpha(1)$ starting from $\alpha(0)$. In this case you can think the elements of $I$ as instants of times and the point $\alpha(t)$ is the position one has at the instant $t$ when moving through the path $\alpha$ (that is a trajectory).

Of course you don't really need of this in order to work with paths nor to prove their properties but it can helps building some intuition and so it can make it easier to work with this stuff.

Now I believe it can help in getting a better understanding about homotopies using this slight different definition: an homotopy is a continuous mapping $H \colon X \times I \to Y$. Note that in this definition we don't make any reference to mappings between $X$ and $Y$.

Using currying you could see also an homotopy as a mapping $H \colon I \to \mathbf {Top}[X,Y]$ (where $\mathbf {Top}[X,Y]$ is the set of continuous functions from $X$ to $Y$).

With this in mind it could really seem appealing to consider homotopy as paths of continuous mappings, that is as continuous functions of type $I \to \mathbf{Top}[X,Y]$.

Unfortunately this is not such a good idea: there are some important complication (for start in order to talk about continuous functions with values in $\mathbf{Top}[X,Y]$ you need to put a topology on the set of continuous functions) which would make really hard to work with a definition of homotopy as path of functions.

Nonetheless it can be helpful to guide intuition by looking at homotopies as wanna be paths between continuous functions, although in proving theorems one has to use the more manageable classical definition.

Hope these remarks can help you build some intuition behind these basic concept of algebraic topology.

p.s.: It is also possible to define homotopies as continuous mappings of the form $X \to \mathbf{Top}[I,Y]$: if you put the compact-open topology on $\mathbf{Top}[I,Y]$ this definition becomes equivalent to the classical one, and this definition can be useful in proving some fact in algebraic topology, nonetheless the classical definition of homotopy is the more useful in practice, at least in my experience.

share|cite|improve this answer
1  
Feel free to ask if you need additional clarifications. – Giorgio Mossa Feb 10 at 14:17

John, I can well understand how frustrating it is to be given a shed load of definitions but no intuitive explanations of what is actually happening. I think that what you really need is to be able to speak to someone face to face about homotopy theory, someone who knows the subject well and is keen to help. I don't really thing that MathsStackExchange will be able to furnish you with this, as you can see in the comments people have already started trying to help, but I don't think these explanations, well intentioned as they are, are really going to get you feeling confident with homotopy theory and I think that they might actually make things more confusing. I have a couple of points to make that I hope will be helpful.

1) Some of the definitions you gave seem a bit dodgy and had stuff missing, who is teaching the course and which uni are you at? You might want to look into getting a textbook on homotopy theory which might be better if you lecture notes aren't very good (This site could well help with recommendations).

2) I would recommend going and asking your lecturer for help or try to find a PhD student working in that area, asking for someone to discuss their subject area with you should be something they are pleased about and this will mean you can have a proper discussion with them about it.

3) Don't give up, you will be able to get on top of this and it may take time and effort but I promise it will be well worth it. Algebraic topology is a beautiful subject area but you're right that definitions aren't useful without intuition and understanding as to what they mean.

4) Try looking at simple examples (again people on here will probably be able to suggest good places to look for these, youtube can be good for this), considering the very simple cases is really important to building intuition since in general the underlying theory is the same for simple or complicated cases, hence understanding it in the simple case will mean you can then apply this understanding in the more complicated case.

share|cite|improve this answer

Ultimately, a homotopy is exactly with the definition says it is. Topology, especially algebraic topology, is a fiddly subject with very precise definitions, and intuition isn't reliable when you look at more exotic spaces. If you're trying to look at specific examples and then generalize, then you're going about it in the wrong direction. Look at the definition, grok it, and then generate your own examples to build up intuition.

To answer your question in brief, a path on a space is a map $f:I \to X$, where $I = [0, 1]$. It's often useful to look at a closed path or loop, which is a path $f$ with $f(0) = f(1)$. In the definition you wrote, think of a homotopy $h:I \times X \to Y$ as a map $h:I \to \operatorname{Map}(X, Y)$, where $\operatorname{Map}(X, Y)$ is the space of continuous maps $X \to Y$. Thus $f, g:X \to Y$ are homotopic iff they're connected by a path in $\operatorname{Map}(X, Y)$.

share|cite|improve this answer

So the first thing that is going on is that somewhere in your lecture notes or in your text book, they defined "map" to mean "continuous function".

Now, I only know this because your quotes won't make sense if that isn't true. So I work backwards from the definitions you are providing me to work out the definitions of the terms used in your definitions. This is sub-optimal, but mathematics is often full of jargon: being able to guess at jargon helps.

A good way to approach this problem is whenever someone uses a term that isn't the totally typical one (ie uses map instead of function), put the words "sufficiently nice" in front of it, and suspend judgement on how nice it needs to be until later. You'll sometimes guess wrong, but it helps offload the work required to understand the definition until later.

And "nice" basically means "has just barely the properties needed to make the definition/theorem true, in a way that is consistent with the other theorems and definitions nearby".

This relies on the fact that mathematicians are lazy, and define away problems when they run into them.


Now, let's apply this guess.

A homotopy between maps $f,g:X \rightarrow Y$ is a map $h: X \times I \rightarrow Y$ such that

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ where $x \in X$ and $I=[0,1]$. We say maps $f,g$ are homotopic.

and it adds

A homotopy deforms the map $f$ continuously to $g$.

Replace map with continuous function:

A homotopy between continuous functions $f,g:X \rightarrow Y$ is a continuous function $h: X \times I \rightarrow Y$ such that

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ where $x \in X$ and $I=[0,1]$. We say continuous functions $f,g$ are homotopic.

and it adds

A homotopy deforms the continuous function $f$ continuously to $g$.

which makes things make a bit more sense.

Next, I'll deconstruct this:

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$

So $h|_{X \times \{0\}} = f$ and $h|_{X \times \{1\}} = g$. $h(?,0)$ is $f(?)$ and $h(?,1)$ is $g(?)$.

(For some function $z$, $z|_{Set}$ is $z$ restricted to $Set$, ie if we only talk about the part of $z$ that is on $Set$, what function do we get? Here I'm lazy again, and despite the fact that $h|_{X \times \{0\}}$ is a function from $X \times \{0\} \rightarrow Y$ and $f$ is a function from $X \rightarrow Y$, I say they are equal, because $X \times \{0\}$ is basically $X$.)

And, most importantly, it is continuous -- so its behavior at the end points restricts (to some degree) what it does in the middle!

The existend of a homotopy means $f$ and $g$ are homotopic. Often we do not care what the homotopy looks like, merely that it exists.

If it exists, then we can "continuously deform" $f$ into $g$ by looking at $h(?,t)$ where $t$ goes from $0$ to $1$. When $t=0$ we get $f$, when $t=1$ we get $g$, and in the middle we get a continuous deformation between the two.

You can think of $h:I \rightarrow (X \rightarrow Y)$, but that requires a friendly definition of continuity on continuous functions from $X \rightarrow Y$, and it might lead to some complications. $h$ in a sense maps $I$ (the unit interval) to a continuous deformation of functions that starts at $f$ and ends at $g$.


Take $X=\{x\}$ the space with a single element $x$. Then a map $f:X \rightarrow Y$ is the same as and element $f(x) \in Y$. A homotopy $h: f \simeq g: X \rightarrow Y $ is the same as a path $h: I \rightarrow Y$ with initial point $h(0)=f(x)$ and terminal point $h(1)=g(x) \in Y$. A homotopy $h: f \simeq f: X \rightarrow Y$ is the same as a closed path $h: I \rightarrow Y$.

Now here we see a mathematican being lazy. If you have the space $I \times \{x\}$, that is basically the same space as $I$, because the cross product of a space with a set containing a single element is basically the same as the space. What the single element is is mostly unimportant.

So they silently and without comment drop the $\{x\}$. Lazy mathematician.

A "path" from a point $a$ to a point $b$ both in $Y$ is just a map $p : I \rightarrow Y$ where $p(0) = a$ and $p(1) = b$. And remember a map is a continuous function.

Each path from England to Sinapore is a different path. Go north 2 km, the strait to England? Yep. Travel in a spiral around the world 5 million times, then stop in England? Yep. Go to the moon? Yep. Travel into a black hole, exit out a white hole and then swing around to England? Yep. Use the cantor set so you only move on a set of measure zero and end up in England? Yep. All (cateogies of) paths to England.

But do remember that we don't have differentiability: just continuity at this point. Paths can be non-differentiable.

A path from 1 to 2 in R might be $f:I \rightarrow R$ such that $f(x) = x+1$. Another is $f(x) = x^2+1$. Or you could use the cantor set to define the function.

In topology, it is the exitence of any path you more often care about, not the definition of a given path. We get theorems about paths existing without having to construct them.

Now, what is worse is that pictures cannot work: they can give trivial examples (like the ones above), but a non-differentiable path cannot be drawn: every drawn line is differentiable.

share|cite|improve this answer

I have just started to watch these video lectures on algebraic topology and the lecturer is excellent and really focused on building up a good understanding of the intuition behind algebraic topology.

Algebraic Topology: a beginner's course - N J Wildberger

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.