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I am learning algebraic topology and reading Lee's book. Here he used the graphs to prove that two paths are homotopic. Although there is some explaination in the book, I can not understand clearly what does the graph mean, why we can draw it like this and how to use these graphs to prove two paths are homotopic. So may I please ask if someone could explain it to me? Thanks.

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2 Answers 2

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One is trying to define a map from $I\times I \to X$. The idea of these pictures is that he decomposes $I \times I$ into certain regions, on which the map is something easy to define. For example, the function that sends a region of $I \times I$ constantly to $p$.

A more complicated example is $f((1-t)s + t(2s - 1))$, which is an interpolation between $f(s)$ and $f(2s - 1)$, the former defined on $[0,1]$ and the latter defined on $[1/2,1]$, going twice as fast.

This function is different from the one defined by the author, and in particular the domain is less nice (the domain is $s \in [t / ( 1+ t),1]$, which you can verify by some algebra), but maybe it brings out more clearly that the construction is an interpolation.

If you vary $t$ and solve for $s$ in $(1 -t ) s + t(2s - 1) = s_0$, you get the curves similar to those the first picture, and you also see the domain. They are supposed to be evocative of the function; it's not accurate that they are level sets, because for all we know $f$ might be constant. They are level sets for the reparametrization formula, in my example $(1 -t ) s + t(2s - 1) = s_0$.

I'm tempted to call $s$ the spatial parameter and $t$ the time parameter, because if $t$ is fixed we imagine a path parametrized with coordinate $s$, and the path evolves as $t$ changes.

Then you use the following lemma to glue together these partially defined functions into a continuous function. (The author calls this the gluing lemma.)

If $X = A \cup B$ with $A$ and $B$ closed, then if we are given continuous functions $f_A : A \to Z$ and $f_B B \to Z$ so that $f_A | _{ A \cap B } = f_B |_{A \cap B}$. Then there is a unique function $f : X \to Z$ that is defined by $f |_ A = f_A$ and $f|_B = f_B$, and moreover it is continuous. (To check continuity, take the preimage of closed sets.)

In this setting, we say that $f_A$ and $f_B$ "glue."

So the idea for building these kind of pictures is that we know what we want to homotopy to look like in certain situations, and so we write down analytic formula inspired by our intuition, and check that they glue using the above lemma. Generally this involves setting up various interpolations.

For the second picture, with words:

The lower triangle consists of the constant path at $f(t)$. The upper left triangle is $f$ running at double speed up to time $t/2$. The upper right triangle is $f$ starting from $f(t)$ and running backwards at double speed.

On each triangle write down an analytic expression for these descriptions, and check that they glue according to the lemma. Then the top gives you f followed by backwards f, and the bottom gives you the constant path, so the square gives you a nullhomotopy of $[f] [f]^{-1}$.

You can visualize this homotopy as follows: At first, we run f followed by f inverse. Then we run $f$ not quite to the end let it wait a bit and run f inverse from where we stopped $f$. We continue on in this way - this is how one might guess the decomposition of $I^2$ as in the second picture.

Hope it helps. I agree this is confusing.

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This is a typical representation of a homotopy.

In particular, The author is asking you to interpret the "time" interval $[0,1]$ as a geometric object.

For example, let $X=[0,1]$ and $Y=0$. We claim that these two spaces are homotopic.

Let $F:X \times [0,1] \to Y$ be given by $F\mid_t(x)=(1-t)x$. At time $0$, this is identity on $x$, and at time $1$, it is identity on $Y$. If you wanted to draw the picture, you would draw the image $F(X \times \{t\})$ for every moment in time and stacked these images , you would get a shape, namely a triangle.

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  • $\begingroup$ I don't think the OP is using Hatcher's book. $\endgroup$
    – Rob Arthan
    Jul 28, 2017 at 21:40
  • $\begingroup$ Oops. You know what they say about assumptions... $\endgroup$ Jul 28, 2017 at 21:41
  • $\begingroup$ No - what do they say about assumptions? $\endgroup$
    – Rob Arthan
    Jul 28, 2017 at 21:43
  • $\begingroup$ Haha. I'm sorry (if you assume you make an ass out of you and me.) I'm not sure why I assumed this was hatcher. $\endgroup$ Jul 28, 2017 at 21:44
  • $\begingroup$ @AM: thanks - hadn't heard that one before. $\endgroup$
    – Rob Arthan
    Jul 28, 2017 at 21:51

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