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.

  • 3
    $\begingroup$ Could you please include the question in this post? So far it just says "It was stated". $\endgroup$ Jan 24, 2020 at 20:04

1 Answer 1


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.

  • $\begingroup$ Ok, so taking the quotient is actually not necesarily a h.e., but being relative does not take into account what happens in there so taking the quotient does not affect the homotopy. Is this what you mean? $\endgroup$
    – astro
    Jan 25, 2020 at 13:40
  • $\begingroup$ @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). $\endgroup$
    – blancket
    Jan 25, 2020 at 13:49
  • $\begingroup$ Thank you, that was exactly what I was missing from the facts. $\endgroup$
    – astro
    Jan 25, 2020 at 14:08

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