2
$\begingroup$

Lets say I have a weak homotopy equivalence $Y \rightarrow Y'$ between arbitrary topological spaces. Does the dashed arrow in

$$\begin{array}{ccc} Y_{CW}&\dashrightarrow&Y'_{CW}\\\ \downarrow&&\downarrow\\\ Y&\rightarrow&Y' \end{array}$$ always exist, such that the diagram commutes? Here $Y_{CW}, Y'_{CW}$ are the corresponding weakly equivalent $CW-$complexes, and the vertical arrows are weak equivalences.

$\endgroup$

1 Answer 1

2
$\begingroup$

There certainly is not typically a dashed arrow that makes the diagram literally commute. For instance, let $Y_{CW}$, $Y'_{CW}$, and $Y$ all be a point, and $Y'=[0,1]$, with $Y\to Y'$ sending the point to $0$ and $Y'_{CW}\to Y'$ sending the point to $1$.

However, there is a dashed arrow making the square commute up to homotopy. You can ignore $Y$ and just consider two weak equivalences $Y_{CW}\to Y'$ and $Y'_{CW}\to Y'$, and you wish to find $Y_{CW}\to Y'_{CW}$ making the diagram commute up to homotopy. This follows from the fact that a weak equivalence induces a bijection on homotopy classes of maps from any CW complex. In particular, the weak equivalence $Y'_{CW}\to Y'$ induces a bijection between homotopy classes of maps $Y_{CW}\to Y'$ and $Y_{CW}\to Y'_{CW}$ (So in fact, you only need to assume that $Y'_{CW}\to Y'$ is a weak equivalence, and that $Y_{CW}$ is a CW complex.)

$\endgroup$
1
  • $\begingroup$ Thanks. I actually meant to prove the statement you mention, when i came upon this question. Ive found the theorem in Hatcher, so im all good now : ) $\endgroup$
    – amueller
    Sep 22, 2017 at 11:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .