(wrong) Proof of stability of homotopy equivalences under pullback

Question

In studying the category Top localized by homotopies, I asked me this question:

"Is homotopy equivalence stable by pullback (base change)?"

I know that it is necessary to have a further condition (pulling along a fibration), but I have a proof in the general case that I cannot disprove (to be honest, there is a passage I didn't finish, which is true if we assume that all the spaces involved are compact). Here it is:

Consider $f: X \to Y$ a homotopy equivalence, $p: Y' \to Y$ a map, and $f':X' \to Y'$ the pulled back map, i.e. $X'$ is the pullback of $X,Y'$ over $Y$ and $f'$ is the resulting projection on $Y'$. We denote by $q: X' \to X$ the other projection. Now recall that:

1. The pullback of a fibration is a fibration;
2. Homotopy equivalences are precisely maps which are both fibrations and cofibrations;
3. The map $A \to Cyl(g)$ is a cofibration and $Cyl(g) \to B$ is an homotopy equivalence, for any $g: A \to B$;
4. If $a$, $ab$ are homotopy equivalences, then $b$ is.

We begin the proof. A) Factor $f:X \to Y$ as $X \to Cyl(f) \to Y$. By (3), $Cyl(f) \to Y$ is a h.e.; by hypothesis, $X \to Y$ is a h.e.; thus by (4) $X \to Cyl(f)$ is a h.e. In particular, by (2) is a fibration.

B) We have a canonical map of diagrams from $X' \to Cyl(f') \to Y'$ to $X \to Cyl(f) \to Y$, because of the natural map $Cyl(f') = Cyl(X') \cup_X' Y' \to Cyl(X) \cup_X Y $. In the next step we show that all the squares in this rectangle are cartesian (I still can't make diagrams in latex; make yourself on your paper to follow the proof please! :) ).

C) The exterior square made up of $X,Y,X',Y'$ is cartesian by definition. Here is the crucial step which I doubt. The square $Cyl(f), Y, Cyl(f'), Y'$ is cartesian; indeed the map $T: Cyl(f') \to Cyl(f) \times_Y Y'$ is explicitly given by

$$ T(x',t) = ( (q(x'),t), f'(x') ) $$ (which are in fact such that $fq(x') = pf'(x')$ ); $$T(y') = ( p(y'), y')$$ Indeed, they agree on the intersection $X' \times \{1\}$: $$T(x',1) = ( (q(x'),1), f'(x') ) = ( f'q(x'), f'(x') ) $$ $$T(f'(x') ) = ( pf'(x'), f'(x') )$$ Which are equal by commutativity of original diagram.

BIJECTIVITY of T: Take $(a,b) \in Cyl(f) \times_Y Y'$. If $a=(x,t) \in X \times I$, the coherence condition says that $f(x) = p(y')$. This defines a unique element $ x' \in X'$ such that $q(x') = x, f'(x') = y'$. In other words, $T(x',t) = (a,b)$. If $a=y \in Y$, the coherence condition says that $a=p(b)$. Thus $T^{-1}(a,b)=\{b\}$.

I have not proved that $T^{-1}$ is continous, but for example this is true if all the involved spaces are compact (e.g. finite CW complexes).

By (2 of 3) principle of cartesian squares, also the square $Y, Cyl(f), X', Cyl(f')$ is cartesian.

D) Here is the main trick. $X' \to Cyl(f')$ is a cofibration by (3), it is a fibration by (1), and thus by (2) is a homotopy equivalence. $Cyl(f') \to Y'$ is always an homotopy equivalence. This yields by composition that $X' \to Y'$ is a homotopy equivalence.

Thank you for your help!

Bonus: if my question is false, it is still true that given maps $ f: X \to Y$, $p:Y' \to Y$ with $f$ homotopy equivalence, then there exist a fourth space $X'$ with maps $q:X' \to X, f': X' \to Y'$ such that the diagram is commutative, $f'$ is a homotopy equivalence, but not necessarily the square is cartesian?

Answer 1

I haven't read through your entire argument, but it's not true that a homotopy equivalence is a fibration/cofibration, or vice versa. For example,

• The map $\{0\} \hookrightarrow \{0, 1\}$ is a fibration and cofibration, but not a homotopy equivalence.
• The map $\{0\} \hookrightarrow I$ is a homotopy equivalence but not a fibration.
• The map $I \to \{0\}$ is a homotopy equivalence but not a cofibration.