0
$\begingroup$

I am struggling immensely with topology since the start of the course, probably due to its extremity; the explanations are either "very rough" or "very strict and rigid and hard to comprehend." Either 0 or 100, no 40 or 50. As a complete newbie, here's a question I am approaching which I managed 20% (if, at all).

The question is with respect to the following diagram:

Commuting maps

Here is the question

For any $f: X \rightarrow S^1$, define $\bar{f}: f^* \mathbb{R} \rightarrow \mathbb{R}$ and $q:f^* \mathbb{R} \rightarrow X$, and $p: \mathbb{R} \rightarrow S^1$ specifically, $t \rightarrow e^{2 \pi i t}$. Prove that for any $x \in X$, the inverse image $q^{-1}(x) \subset f^* \mathbb{R}$ is homeomorphic to $\mathbb{Z}$ with the discrete topology. $f^*\mathbb{R}$ is the pullback of $p$ on $f$, defined $f^*\mathbb{R}= \{(x,t) \in X \times \mathbb{R} : f(x) = p(t) \in S^1\}$

1.$\,$My remark about topology's extremity; homeomorphism. Sure, I get the "vague" idea that it is a type of mapping and just as how say coffee cups can be transformed into a torus but not a sheet of paper, homeomorphic (topological) spaces are "equivalent." That is, a "homeomorphism exists" between them. But specifically, what are they? My notes gives me the example $[0,1)$ and $(0,1]$ are homeomorphic with $x \rightarrow 1-x$.

a) I thought homemorphisms are defined in "topological" spaces, $[0,1)$ and $(0,1]$ seem to me just "sets." What is the topology defined on them?

b)Sure $x \rightarrow 1-x$ is a "map" but I mean, how does this relate to the "intuitive rough explanation" of transforming a torus and a coffee cup? $[0,1)$ is just an interval, a straight line so... a straight line transformed to a straight line? Is that what's going on here?

Then the stuff gets very rigid; homeomorphisms are bijective continuous maps. Well, I can perhaps try to ram the definition down my throat without a great understanding of it, but while people told me "topology is pretty much about drawing stuff out" I can't imagine anything as to what "homeomorphism" is.

Okay, back to the question, what I figured so far

$(\mathbb{Z}, \tau _{disc})$, $\tau_{disc}$ is the power set, so essentially, this topological space has elements which are all open(right?). So, the homeomorphism between the subset $Q=\{q^{-1}(x): x \in X\}$ should have its preimage i.e. Q to be an open set itself(by definition of "continuity" of a map for topological spaces, which a homeomorphism must satisfy).

is all. I get stuck here; I mean, I don't know what $X$ is, I don't know what topology is defined on it(is there? Or do I have the liberty of coming up with whatever that's convenient?) and I have no means of deducing that $Q$ is open. Well, can I define a discrete topology on $Q$ too? That should make every element or subset open in $Q$.

And even if I manage to show "continuity" I don't know how "bijectivity" can be shown.

The issue is, I really have no means of finding a "homeomorphism" in general. How do people show these? A good guess? Intuition? Years of practice? I don't immediately see "hmm, ah, a map like this might work...or if not, probably this one..yeah that works"

Would someone please show me how this can be solved? Thank you so much in advance

$\endgroup$
  • $\begingroup$ What is $f^\ast\Bbb R$? $\endgroup$ – Pedro Tamaroff Feb 2 '16 at 21:35
  • $\begingroup$ Hi just added the definition to it. $\endgroup$ – John Trail Feb 2 '16 at 21:36
  • 1
    $\begingroup$ A few remarks. 1) Intuitively, a homeomorphism is like a "continuous deformation". For example, you can deform a half-sphere into a circle. And indeed, they are homeomorphic. a) $[0,1)$ and $(0,1]$ have a natural topology: the topology generated by intersections of open intervals $(a,b)$ with those sets. Such intersections might not be open intervals, because if $0\in(a,b)$, $0\in(a,b)\cap[0,1)$, which is thus $[0,\max\{1,b\})$. b) $x\mapsto1-x$ is like taking a line segment and flipping it and superimposing it with itself. Might not be what you thought of with "deformation" in 1), but it is… $\endgroup$ – MickG Feb 2 '16 at 21:44
  • $\begingroup$ … still a homeomorphism. In general, if you deform continuously and bijectively, or translate, or rotate, you are likely to get a homeomorphism. $\endgroup$ – MickG Feb 2 '16 at 21:45
  • $\begingroup$ Of course, be careful with "deformations": contracting a circle to a point is certainly a continuous deformation, but definitely not a homeomorphism. $\endgroup$ – MickG Feb 2 '16 at 21:46
2
$\begingroup$

It seems to me that you have all the correct ideas about topology. Topology works best as a mixture of intuition and precision. The fuzzy intuitive dreaming is fun, but it needs to interact correctly with the precise and rigid part. For instance, dreamy intuition can be a helpful step in formulating a precise mathematical description of what is true and what is not true. On the other hand, precise calculation might be helpful in dreaming up the correct picture.

In your penultimate paragraph where you write "...I really have no means of finding a homeomorphism...", you then go on to give a precisely correct description of how one proves that two spaces are homeomorphic: you must somehow dream up an appropriate formula for a homeomorphism, write that formula down precisely, and then prove that the formula you wrote down satisfies the definition of a homeomorphism. Keep in mind, the property that two spaces $X,Y$ are homeomorphic is an existential property:

  • $X$ and $Y$ are homeomorphic if there exists a homeomorphism $h:X \to Y$.

Like any existential property, there are often (usually) no shortcuts to dreaming up a precise description of the appropriate object whose existence is being asserted, and then proving that this object satisfies all the properties needed. But you might have to do some precise work to aid the dreaming up stage.

For the mathematical question you asked, before you can "see" or "dream up" or "intuit" a correct formula for a homeomeomorphism from the space $q^{-1}(x)$ to the space $\mathbb{Z}$, you first need to very carefully pick apart the meaning of the expression $q^{-1}(x)$ for $x \in X$. To do this you should follow your nose through the trail of definitions, aided by your background knowledge of complex numbers, trigonometry, etc.

  • First, apply $f$ to $x$, and you get $f(x) \in S^1$. Give the point a name, perhaps $z=f(x)$.
  • Next, use the definition of $S^1$ to write an expression for $z$. For this problem, the most convenient and appropriate expression might be $$z = f(x) = \cos(2 \pi t_0) + i \sin(2 \pi t_0) = e^{2 \pi it_0}, \quad t_0 \in [0,1) $$ The reason this is the most convenient and appropriate expression is because it matches up with the description of the map $p$. The number $t_0$ is of course a well-defined number depending only on $z$.
  • Next, use the definition of $p$ and your knowledge of trigonometry etc. to write an expression for $p^{-1}(z)$: $$p^{-1}(z) = \{t_0 + 2 \pi n \, \bigm| \, n \in \mathbb{Z}\} $$
  • Next, use the definitions of $f^*\mathbb{R}$, of $\bar f$, and of $q$, to show that $$q^{-1}(x) = \{(x,t) \, \bigm| \, t \in p^{-1}(z)\} = \{(x,t_0 + 2 \pi n) \, \bigm| \, n \in \mathbb{Z} \} $$

Now can you write down a formula for a homeomorphism $q^{-1}(x) \mapsto \mathbb{Z}$?

$\endgroup$
  • $\begingroup$ Thanks for all the help, so now I need some map $(x,t) \rightarrow k$ where $k \in \mathbb{Z}$ yes? My issue is, because I don't know what $X$ is(real?complex?integers? rational? irrational?) I need to come up with a map that regardless of whatever the elements of $x$ is, it will take $(x,t)$ and spit out an integer... yes? Does something like a step function(maybe floor or ceiling function) qualify as a map..? But then I guess it's not bijective... $\endgroup$ – John Trail Feb 3 '16 at 17:40
  • 1
    $\begingroup$ @JohnTrail notice how all points of $q^{-1}(x)$ have $x$ as their first component and a number as the second. You can very well neglect the first component: $q^{-1}(x)$ is clearly in bijection with the set $p^{-1}(z)$ via the projection onto the second component, and I claim this map is a homeomorphism if on $q^{-1}(x)$ we place the topology induced on it by the product topology on $f^\ast\mathbb R$ and on $p^{-1}(z)$ that induced by the euclidean topology on $\mathbb R$. Which btw are both discrete topologies. Can you see a way of extracting an integer from an element of $p^{-1}(z)$,… $\endgroup$ – MickG Feb 3 '16 at 22:23
  • 1
    $\begingroup$ …directly from the way Lee has written the elements of that set? HINT: they depend on an integer parameter… $\endgroup$ – MickG Feb 3 '16 at 22:24
  • $\begingroup$ Following up on the further hint of @MickG, what you need is some map $(x,t_0+2\pi n) \mapsto k$ where $k \in \mathbb{Z}$. $\endgroup$ – Lee Mosher Feb 4 '16 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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