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Show that $f:X \rightarrow Y$ is a homotopy equivalence if there exist maps $g,h:Y \rightarrow X$ such that $fg \simeq \mathbb{1}$ and $hf \simeq \mathbb{1}$.

Why isn't this trivial. Surely if f is a homotopy equivalence we get the maps for free with say g=h.

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You are assuming you have these maps, not that you have a homotopy equivalence. The problem then is that you want to conclude that you can take $h=g$, which is not a priori obvious. –  Alex Youcis Mar 4 '12 at 19:46
    
To add on Alex's comment, the implication is IF exists $h,g$ etc. THEN $f$ is homotopy equivalence. What you are asking is the other way around, you assume that $f$ is homotopy equivalence and take $g=h$; you need to prove the other direction. –  Asaf Karagila Mar 4 '12 at 19:49
    
@AsafKaragila It still seems really trivial. –  danielr1234 Mar 4 '12 at 19:52
    
So you have g,h and identity. Then surely, by definition f is a homotopy. –  danielr1234 Mar 4 '12 at 19:53
    
No one said that Chapter 0 exercises are difficult, most times these exercises are meant to let you play with the definition a bit and try some easy things so you can slowly wade into the material later on. –  Asaf Karagila Mar 4 '12 at 19:53
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2 Answers 2

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This answer assumes the following theorem, so if it's not something you feel you can use you'll need to prove it first:

If $f_1,g_1\colon X\to Y$ are homotopic, and $f_2,g_2\colon Y\to Z$ are homotopic, then the compositions $f_2\circ f_1$ and $g_2\circ g_1$ are also homotopic.

Assuming this theorem, we show that if $fg\backsimeq1$ and $hf\backsimeq1$ then $g\backsimeq h$, so $1\backsimeq fg\backsimeq fh$ and $f$ is by definition a homotopy equivalence (with inverse homotopy equivalence $h$). To do this, we use the above theorem and the associativity of composition to find:

$$h=h\circ 1\backsimeq h(fg)=(hf)g\backsimeq 1\circ g=g$$

So $g\backsimeq h$ and we have the result as above.

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How do you prove the thing you are using. Hatcher asks to prove something like that earlier. –  danielr1234 Mar 4 '12 at 20:21
    
Reasonable question - does Hatcher not prove it himself? It's a fairly important theorem for a lot of the rest of the book (assuming you're talking about Algebraic Topology and not some other Hatcher book). If I recall correctly, you can prove the result I mentioned directly, by taking a homotopy $H_1$ from $f_1$ to $g_1$, and a homotopy $H_2$ from $f_2$ to $g_2$, and "composing them" to get the homotopy you want. The inverted commas are because you'll need to fiddle with $H_1$ a bit (i.e. make it map to $Y\times I$ instead of just $Y$) so that $H_2\circ H_1$ is defined. –  Matt Pressland Mar 4 '12 at 20:37
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If you are familiar with categories then this can help:

$f:X\rightarrow Y$ is a homotopy equivalence in category $\mathbf{Top}$ is exactly the same statement as: $\left[f\right]:X\rightarrow Y$ is an isomorphism in category $\mathbf{hTop}$.

$g,h\in\mathbf{Top}\left(Y,X\right)$ with $fg\simeq1$ and $hf\simeq1$ is exactly the same statement as: $\left[g\right],\left[h\right]\in\mathbf{hTop}\left(Y,X\right)$ with $\left[f\right]\left[g\right]=1$ and $\left[h\right]\left[f\right]=1$.

Based on the last result we find: $\left[g\right]=\left[1\right]\left[g\right]=\left(\left[h\right]\left[f\right]\right)\left[g\right]=\left[h\right]\left(\left[f\right]\left[g\right]\right)=\left[h\right]\left[1\right]=\left[h\right]$.

Then we have $\left[f\right]\left[g\right]=1$ and $\left[g\right]\left[f\right]=\left[h\right]\left[f\right]=1$ or equivalently: $\left[f\right]$ is an isomorphism.

The last statement is exactly the same statement as: $f$ is a homotopy equivalence.

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+1 I like your categorical point of view –  magma Sep 27 '13 at 8:04
    
Ja, dat was me al duidelijk (= that was allready clear to me).:) –  drhab Sep 27 '13 at 8:19
    
@magma Thanks for edits. Oh, I am so clumsy... –  drhab Sep 27 '13 at 8:24
    
You are welcome. I have't received your email yet :-) –  magma Sep 27 '13 at 8:41
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