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Reading Allen Hatchers book (available online via this link) on Algebraic Topology, it states on page 3 that homotopy type defines an equivalence relation. The symmetry and reflexiveness are immediately seen, but the transitivity requires a little work. So my question is basically if this is the way to do it:

My attempt

Assume $X\simeq Y$ and $Y\simeq Z$. We must show that $X\simeq Z$. We have functions $$ \begin{align} X\overset{f_1}{\longrightarrow}&Y\overset{f_2}{\longrightarrow}Z\\ X\overset{g_1}{\longleftarrow}&Y\overset{g_2}{\longleftarrow}Z \end{align} $$ where $f_1g_1,g_1f_1,f_2g_2$ and $g_2f_2$ are homotopic to identities in the relevant spaces.

I claim that $f=f_2f_1$ and $g=g_1g_2$ are homotopy equivalence maps for $X\simeq Z$. To see this consider: $$ H(z,t)= \begin{cases} f_2\circ H_1(y,2t)\circ g_2,&\mbox{ for }t\in[0,0.5]\\ \quad\\ \quad\\ H_2(z,2t-1),&\mbox{ for }t\in[0.5,1] \end{cases} $$ where $H_1$ is a homotopy of $f_1g_1$ and $\mathbb{1}_Y$ whereas $H_2$ is a homotopy of $f_2g_2$ and $\mathbb{1}_Z$. Then $H(z,0)=fg$ and $H(z,1)=z$ so that $H$ is a homotopy of $fg$ and $\mathbb{1}_Z$. Of course one should check that $H$ depends continuously on $t$ especially for $t=0.5$ where $H(z,0.5)=f_2g_2$ by both definitions which will make that condition hold.

By very similar arguments it can be established that $gf\simeq\mathbb{1}_X$ and we are done.

Does this make sense and is it the way it is supposed to be carried out?

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  • $\begingroup$ The symmetry and reflexivenes ARE immediately seen... wait reflexivenes isn't a word $\endgroup$ – Squirtle Feb 10 '14 at 23:24
  • $\begingroup$ @Squirtle: Sorry, cannot count to two :) $\endgroup$ – String Feb 10 '14 at 23:26
  • $\begingroup$ @Squirtle: with an extra 's' it becomes one... $\endgroup$ – String Feb 10 '14 at 23:27
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This is how it is supposed to be done, yes. You have to go from $0$ to $1$, and by going through the first mapping at twice the speed then going through the second mapping at twice the speed you get the proper speed for the whole composition of maps. (that way you spend half the total time on each)

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  • $\begingroup$ Very nice, than you for your time! $\endgroup$ – String Feb 10 '14 at 23:28
  • $\begingroup$ My spelling has gone mad :S. BTW I am NOT a native speaker - isn't that a valid excuse? $\endgroup$ – String Feb 10 '14 at 23:34
  • $\begingroup$ lol.... its fine. I am feeling very grammatical because my C* algebra professor today humorously wanted to point out the difference of finite-dimensional and finite dimensional depending on their location and function in a sentence. $\endgroup$ – Squirtle Feb 10 '14 at 23:45
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    $\begingroup$ Change half to twice! $\endgroup$ – PVAL-inactive Feb 11 '14 at 1:08
  • $\begingroup$ thanks..... edit reflects this $\endgroup$ – Squirtle Feb 11 '14 at 18:19
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Once you have shown that if $f\simeq f'$ and $g\simeq g'$, also $gf\simeq g'f'$, you can write $$f_2f_1g_1g_2=f_2(f_1g_1)g_2\simeq f_2 1_Y g_2\simeq 1_Z$$ More generally, the property stated above can be expressed as "composition of homotopy classes is well defined by the formula $[g][f]=[gf]$". This means that the topological spaces with the classes of maps form a category $h\mathbf{Top}$, and two spaces are isomorphic in this category if they are homotopy equivalent. But "isomorphic to" is an equivalence relation.

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