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Brayton Gray's book Homotopy Theory is the only book I've seen that states and proves the following theorem.

Lemma 16.17. Suppose $f: X \to Y$ is a base point preserving map. $f$ is a weak homotopy equivalence iff given any CW pair $(L, L_0)$ and maps $\alpha: L_0 \to X$, $\beta: L \to Y$ with $f \alpha =\beta|_{L_0}$ there is a map $g: L \to X$ with $g|_{L_0} = \alpha$ and $fg \simeq \beta (\text{rel } L_0)$.

The proof in the stated book goes through numerous previous theorems, lemmas and problems, which are unnecessarily general and complicated. Is there any other (more direct) proof of this fact? Where can I look for it?

Jesus
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    First, replace $Y$ by $M_f$ to assume that $f$ is an inclusion WLOG. The compression lemma/criterion basically states that $f$ is a weak equivalence iff your criterion holds for $(L,L_0)=(D^n,S^{n-1})$, so sufficiency is clear and necessity remains. The basic idea is to construct $g$ by induction over the skeleta of $L$. Take the characteristic map of a cell, post-compose with $\beta$, use the HEP to homotope it to a map extending the given one on the boundary and then the compression lemma to homotope it rel boundary to a map into $X$. Finally, pass to the limit. – Thorgott Mar 27 '23 at 22:30
  • @Thorgott thanks, this is a great comment. Would you consider writing it as an answer with perhaps a bit more details? – Jesus Apr 05 '23 at 12:54

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Let us assume that $f\colon X\rightarrow Y$ is a map (there is no need for basepoints). It is to be shown that $f$ is a weak homotopy equivalence if and only if for every CW pair $(K,L)$, maps $\alpha\colon K\rightarrow Y$ and $\beta\colon L\rightarrow X$ such that $f\beta=\alpha$, there is a map $\alpha^{\prime}\colon K\rightarrow X$ such that $\alpha^{\prime}\vert_L=\beta$ and $f\alpha^{\prime}\simeq\alpha$ rel $L$. I leave it as an exercise to check that we may replace $Y$ by the mapping cylinder $M_f$ without changing either hypothesis, so that we can assume $f$ to be an inclusion WLOG. Then, $f$ is a weak equivalence if and only if $\pi_n(Y,X,x_0)=0$ for all $x_0\in X$ and $n\ge0$. I leave it as another exercise to check that this is the case if and only if the second property is satisfied for the CW pairs $(K,L)=(D^n,S^{n-1})$, $n\ge0$ (in the case $n=0$, we take $S^{-1}=\emptyset$). This equivalence can also be argued directly, but I find it more convenient to reduce to the case where $f$ is an inclusion anyway. It is thus clear that the second property is sufficient and it remains to argue that it is necessary.

Let me argue first that if $\alpha$ maps the $n$-skeleton of $K^n$ into $X$, then we can homotope it rel $K^n\cup L$ to a map taking the $(n+1)$-skeleton $K^{n+1}$ into $X$. Indeed, for an $(n+1)$-cell $\Phi\colon(D^{n+1},S^n)\rightarrow(K^{n+1},K^n)$ that is not in $L$, we can consider $\alpha\circ\Phi\colon(D^n,S^{n-1})\rightarrow(Y,X)$ and so the $(D^{n+1},S^n)$ case implies that we can homotope $\alpha\circ\Phi$ rel $S^{n-1}$ to a map into $X$. Together with the constant homotopy on $K^n\cup L$, these glue to a homotopy of $\alpha\vert_{K^{n+1}\cup L}$ rel $K^n\cup L$ to a map into $X$. Since the CW pair $(K,K^{n+1}\cup L)$ is cofibered, this extends to a homotopy of $\alpha$ rel $K^n\cup L$ to a map taking $K^{n+1}$ into $X$, as desired.

Now, we can start with the empty case $n=-1$, and homotope $\alpha$ rel $L$ to a map taking $K^0$ into $X$ in the $t$-interval $[0,1/2]$. Then, homotope the resulting map rel $K^0\cup L$ to a map taking $K^1$ into $X$ in the $t$-interval $[1/2,3/4]$. Continue this process inductively. Note that, for each $n\ge0$, the homotopies stay constant on $K^n$ after finitely many steps, so this extends to a well-defined homotopy rel $L$ whose value at $t=1$ is a map taking $K$ into $X$, as desired.

Two textbook references containing closely related results are Theorem 11.12. in Bredon's Topology and Geometry and Lemma 4.6. in Hatcher's Algebraic Topology.

Thorgott
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