$\require{AMScd}$
On page 398 Hatcher states that the Puppe sequence has a naturality property, namely that a map $f \colon (X,A)\to (Y,B)$ of CW-pairs induces maps between the Puppe sequences of these pairs, with homotopy-commutative squares, i.e. the following diagram commutes up to homotopy:
\begin{CD}
A @>>> X @>>> X/A @>\varphi>> \Sigma A @>>> \Sigma X @>>> \Sigma(X/A) @>>> \Sigma^2A @>>> \cdots\\
@Vf|_AVV @VfVV @V\overline{f}VV @V\Sigma(f|_A)VV @V\Sigma(f)VV @V\Sigma(\overline{f})VV @V\Sigma^2(f|_A)VV \\
B @>>> Y @>>> Y/B @>>\psi> \Sigma B @>>> \Sigma Y @>>> \Sigma(Y/B) @>>> \Sigma^2B @>>> \cdots\\
\end{CD}
Here $\Sigma X$ denotes the reduced suspension of $X$ and the map $\overline{f}$ is given by $\overline{f}([x]) = [f(x)]$.
I understand how Hatcher constructed the map $\varphi$ (and $\psi$ similarly), by defining it to be the composition
$$
X/A \to X \cup CA \hookrightarrow (X \cup CA) \cup CX \twoheadrightarrow SA \twoheadrightarrow \Sigma A,
$$
where the first of these maps is a homotopy inverse of the quotient map $X \cup CA \twoheadrightarrow X/A$ and $SA$ denotes the unreduced suspension. Moreover, I see that the first two squares in the diagram commute. I only fail to see why the third square commutes up to homotopy. I have tried showing by hand that $\Sigma(f|_A) \circ \varphi$ is homotopic to $\psi \circ \overline{f}$, but this hasn't led to anything, unfortunately.
Can anyone help me out? Thanks!
Best Answer
Clearly the sequence $$X \cup CA \hookrightarrow (X \cup CA) \cup CX \twoheadrightarrow SA \twoheadrightarrow \Sigma A $$ is natural. Moreover the quotient map $$p_{(X,A)} : X \cup CA \twoheadrightarrow X/A$$ is natural. In fact, if $f : (X,A) \to (Y,B)$, then $\bar f \circ p_{(X,A)} = p_{(Y,B)} \circ f'$, where $f' : C \cup CA \to Y \cup CB$ is the obvious map induced by $f$.
Letting $h_{(X,A)}$ and $h_{(Y,B)}$ be homotopy inverses for $p_{(X,A)}$ and $p_{(Y,B)}$, we get $$h_{(Y,B)} \circ \bar f \simeq h_{(Y,B)} \circ \bar f \circ p_{(X,A)} \circ h_{(X,A)} = h_{(Y,B)} \circ p_{(Y,B)} \circ f' \circ h_{(X,A)} \simeq f' \circ h_{(X,A)} .$$ This shows that the third square commutes up to homotopy.