# Allen Hatcher algebraic topology proof of theorem 4.5 Whitehead

It was stated as the last answer in this post:

Idea behind the proof of Whitehead's Theorem and Compression Lemma

I copy here what is asked there:

I am reading as well the proof of Whitehead's theorem in AHAT and trying to grasp the idea of what happens in the general case when $$f$$ is not cellular without using cellular approximation.

The final line of the argument ''$$( X \times I \sqcup Y, X \times \partial I \sqcup Y) \rightarrow (M_f, X \cup Y) \rightarrow (M_f, X)$$ gives a deformation retraction of $$M_f$$ onto $$X$$'' is, if I understand right, quotient by $$(x,1) \sim f(x)$$ in the first arrow and the homotopy $$g$$ explained in the book in the second. Clearly, this composition would give the deformation retraction we seek if taking the quotient in the first arrow would give a homotopy equivalence, which I am not so sure why is so.. this may be an easy algebraic fact but I'm not on algebra so if it is a silly questions I excuse myself from now lol, anyway, Thank you very much in advance for any help on this matter.

• Could you please include the question in this post? So far it just says "It was stated". Jan 24, 2020 at 20:04

Applying the compression lemma to the inclusion $$(X \cup Y,X) \to (M_f,X)$$, we obtain a homotopy rel $$X$$ from $$X \cup Y \to M_f$$ to a map $$X\cup Y \to X$$. Since $$(M_f,X \cup Y)$$ has the homotopy extension property, this homotopy extends to a homotopy rel $$X$$ from $$\text{id}_{M_f}$$ to a map $$g:M_f \to M_f$$ such that $$g(X\cup Y)\subset X$$. Now apply the compression lemma again to the composition
$$g \circ q: (X\times I \amalg Y, X\times \partial I \amalg Y) \to (M_f, X\cup Y) \to (M_f, X)$$
where $$q: X\times I \amalg Y \to M_f$$ is the standard quotient map. Then we obtain a homotopy $$F: (X\times I \amalg Y)\times I \to M_f$$ rel $$X\times \partial I \amalg Y$$ from $$g \circ q$$ to a map $$X\times I \amalg Y \to X$$. Since the homotopy $$F$$ is "rel $$X\times \partial I \amalg Y$$", it passes to the quotient and induces a homotopy $$\bar{F}:M_f \times I \to M_f$$ rel $$X\cup Y$$ from $$g$$ to a map $$h:M_f \to X$$. Therefore, we have $$\text{id}_{M_f} \simeq g \simeq h$$ rel $$X$$, and hence $$X$$ is a deformation retract of $$M_f$$, as desired.
• @Astro Yes, the map $q$ is not necessarily a homotopy equivalence. The significant fact is that $X\times \partial I \amalg Y$ is fixed during $F$, so $F$ can pass to the quotient (by the universal property of quotient spaces). Jan 25, 2020 at 13:49