In preparation for an upcoming exam, I’m working through old exercises.
Let $X$ be a CW complex with skeletons $X^0 ⊂ X^1 ⊂ … ⊂ X$. Show that for any open $U ⊂ X^{n-1}$ there is an open $V ⊂ X^{n}$ such that $U ⊂ V$ and the inclusion $U → V$ is a homotopy equivalence.
I’m using a pushout definition of CW complexes:
- $X$ has the final topology with respect to all inclusions $X^n ⊂ X$.
- For every $n$ there is a set $I_n$, indexing $n$-disks, so that we have a pushout square: \begin{align*} \require{AMScd} \begin{CD} \coprod_\smash{I_n} S^{n-1} @>{⊂}>> \coprod_\smash{I_n} D^n\\ @V{φ}VV @VV{φ}V \\ X^{n-1} @>>{⊂}> X^n \end{CD} \end{align*}
Okay, what I have done and know:
- $\hat U := φ^{-1} (U)$ is open in $\coprod S^{n-1}$.
- $\hat V := \{x ∈ \coprod D^n\setminus \{0\}; \frac{x}{\lVert x\rVert} ∈ \hat U\}$ is open in $\coprod D^n$.
- $\hat U ⊂ \hat V$ is a deformation retract.
Set $V := U ∪ φ(V)$. I feel like, if I roll up my sleeves, I’m able to prove that
- $V ∩ X^{n-1} = U$, and
- $φ^{-1}(V) = \hat V$,
so that $V ⊂ X^n$ is indeed open. (Although I’d like to see hints for how to do that with the categorical notion of a pushout, in the same spirit of what I’ve done here.)
But how can I show that the inclusion $U → V$ is indeed a homotopy equivalence?
I want to use the deformation retract $\hat U → \hat V$, of course, but I don’t see how I can really push it down to $U → V$.