The question is fairly general. Suppose I have a homotopy commutative diagram of the form

\begin{equation} \require{AMScd} \begin{CD} A @>{f}>> B\\ @V{h}VV @V{i}VV \\ C @>{g}>> D \end{CD} \end{equation}

meaning that there is a homotopy from $i\cdot f$ to $g\cdot h$. Suppose further I'm given a space $A'$ which is homotopy equivalent to $A$, i.e. I'm given maps $a:A\to A'$, $a':A'\to A$ such that $a\cdot a'$ and $a'\cdot a$ are homotopic to the identity, respectively on $A'$ and $A$.

It would seem quite natural for me to have homotopy commutativity for the diagram

\begin{equation} \require{AMScd} \begin{CD} A' @>{f\cdot a'}>> B\\ @V{h\cdot a'}VV @V{i}VV \\ C @>{g}>> D \end{CD} \end{equation}

since the original diagram is commuting up to homotopy, and somehow the spaces are indistinguishable, again up to homotopy. However, I don't see how to prove this: these homotopy equivalences seem to get handy just when they come in pairs one after the other, and here I have just one - namely, $a'$.

The initial problem was related to a similar diagram, where $B=A$, and it seems to hold that

\begin{equation} \require{AMScd} \begin{CD} A' @>{a\cdot f\cdot a'}>> A'\\ @V{h\cdot a'}VV @V{i\cdot a'}VV \\ C @>{g}>> D \end{CD} \end{equation}

homotopy commutes. This $f$ stuck in the middle irritates me, and, although the solution seems to be at hand, I can't seem to find it.

Which is why any hint is greatly appreciated. :-)

  • $\begingroup$ In your first diagram you have defined $h$ as being a map from $A$ to $C$. However in your second diagram $h$ is a map from $A'$ to $C$. Shouldn't this be $h\circ a'$? $\endgroup$ – Jorik Dec 2 '15 at 21:50
  • $\begingroup$ correct! I'm fixing it now, thanks uh also later on $\endgroup$ – nelv Dec 2 '15 at 21:54
  • $\begingroup$ You mentioned something about pairs of homotopies. Do you mean that fact that when $f_1,f_2:X\to Y$ are homotopic and $g_1,g_2:Y\to Z$ are homotopic then $g_1\circ f_1$ is homotopic to $g_2\circ f_2$? $\endgroup$ – Jorik Dec 2 '15 at 21:57
  • $\begingroup$ Actually, I was referring to the pair of homotopy equivalencies $a$ and $a'$, that are homotopic to the identity when composed - but this composition never seem to occur in considering the problem $\endgroup$ – nelv Dec 2 '15 at 22:00
  • $\begingroup$ That actually does occur in your second case now that you've fixed the diagram. You can consider my previous comment a hint, but if you want some more elaboration I could give some. $\endgroup$ – Jorik Dec 2 '15 at 22:02

Suppose $i \circ f \simeq g \circ h$, then you can pick a concrete homotopy $H: I×A → D$ where $I$ is the unit interval and $H_0 = i \circ f$, $H_1 = g \circ h$. (Here $H_t(x)$ denotes $H(t, x)$ and $H_t : A → D$.)

You can precompose this map with $id_I × a'$ to get $\overline{H} : I × A' → D$ defined as $\overline{H}(t\in I, x\in A') = H(t, a'(x) \in A)$. Then the restriction $\overline{H}_0 = i \circ (f \circ a')$ and $\overline{H}_1 = g \circ (h \circ a')$. This is a homotopy which shows $i \circ (f \circ a') \simeq g \circ (h \circ a')$.

Note that you don't use “$a'$ is a homotopy equivalence” in this proof.

  • 1
    $\begingroup$ thank you! a posteriori, it seems easy - I guess that's quite often the case $\endgroup$ – nelv Dec 2 '15 at 22:10

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