# Problem with understanding proof of Van Kampen's theorem

I'm currently reading J.P May's book, "A Concise Course in Algebraic Topology".

I don't understand his proof of the fundamental groupoid version of Van Kampen's theorem, particularly the part where he proves that $\tilde\eta ([f])=\tilde\eta[g]$, if $[f]=[g].$

Let $f$ & $g$ be paths form $x$ to $y$ in $X$. We wish to define $\tilde\eta:\Pi(X)\to\mathscr{C}$,given $\eta_U:\Pi(U)\to \mathscr{C}$.

The book defines $\tilde{\eta}(x)=\eta_U(x)$, and $\tilde\eta([f])=\eta([f])$ if $f$ lies within $U$.If $f$ doesn't lie within $U$, it is the composite path of paths $f_i$ all lying within some $U_i$ and $\tilde \eta([f])=\Pi\eta_{U_i}([f_i])$. He now wishes to show that this is well defined. He says that if $[f]=[g]$, there is a homotopy, $h:f\simeq g$. You can divide the square into a grid of finitely many subsquares, such that $h$ is mapped into one of the $U$ of the cover. He then says that we can choose a grid that refines the partition of $I\times\{0\}$ for $f$ and $I\times\{1\}$ for $g$. Now he concludes that $\tilde\eta([f])=\eta([f])$, thanks to it being the consequence of finitely may relation which hold in one of the $U$. I don't understand the last step.

• Write down the part that you don't understand and ask a more concrete question? Commented Oct 20, 2014 at 14:10
• Sorry. I am editing the question. Commented Oct 20, 2014 at 14:11

It might be useful to compare the proof given in May's book to, say, the more concrete proof of the classical van Kampen in, say, Hatcher's book. The proof given is fairly straightforward, except that it's dressed up in the language of category theory (which is not a bad thing at all, especially in a field like algebraic topology that actively uses the machinery of category theory). The last part of the proof is saying that we can decompose the interval $I\times I$ into a series of small strips of the form $J_i\times I$ with the images of $f\vert J_i$ and $g\vert J_i$ lying entirely in some element $U$ of the given cover ${\scr O} = \{U\}$. On each such strip $J_i\times I$, we have a homotopy $f_i \simeq g_i$. If you trace back through the definition of $\tilde \eta$, that gives the required equality $\tilde \eta[f] = \tilde \eta[g]$. (Specifically, the definition I'm referring to is the statement that "Any path $f$ is the composite of finitely many paths $f_i$, each of which does lie in a single $U$, and we must define $\tilde \eta[f]$ to be the omposite of the $\tilde \eta[f_i] = \eta([f_i])$. Everything after that is just showing that that construction is well-defined.)

• It might be worth mentioning that $f_i \simeq g_i$ is not a homotopy rel basepoints. But when we put them all together, it becomes a homotopy rel basepoints. (If I'm understanding correctly.) Also might be worth mentioning that the reason we can do this pasting is that it is a locally finite closed cover. Commented Apr 16, 2016 at 12:09
• I don't think that May is decomposing $I\times I$ into vertical strips as you mention. He talks of subdividing the square into "subsquares".
– Atom
Commented Oct 24, 2023 at 17:31

This is not an answer, but I had trouble posting it as a comment.

Here's a special case that already seems problematic: let $f : [0,1] \to [0,1]$ be monotone increasing, fixing 0 and 1. $f \simeq \mathrm{id}_{[0,1]}$. Now if $\mathscr{O}$ is a open cover of $[0,1]$, can you find a sequence of homotopies between $f$ and $\mathrm{id}_{[0,1]}$ such that each of them fixes the complement of some $U \in \mathscr{O}$?

This seems to be a lemma you need in order to reparametrise paths.

EDIT: It might be worth comparing May's proof to what Ronnie Brown does in his book. I seem to recall, he avoids some of the issues that come with reparametrisation by using Moore loops