I want to prove that Every path connected CW complex $X$ is homotopy equivalent to a complex $Y$ with a single $0$-cell. Notice that I'm making no assumption on the dimension of $X$ or the number of cells.

My idea would be to take the $0$-cells in $X$ and collapse them to a point. So, I consider $X^{(1)}$ (the $1$-skeleton) and a maximal tree $T$ (maximal among the contractable $1$-dimensional subcomplexes of $X^{(1)}$) in $X^{(1)}$, then $T$ is a subcomplex of $X^{(1)}$ which is a subcomplex of $X$, so $T$ is a subcomplex of $X$. Moreover, $T$ is contractable and contains every $0$-cell of $X^{(1)}$ (so every $0$-cell of $X$). We conclude $X/T$ is a CW complex with a single $0$-cell and the projection $X\rightarrow X/T$ is a homotopy equivalence. The existence of a maximal tree is proved with Zorn's lemma.

Is there some problem with this argument?


1 Answer 1


Your proof is correct.

Here is another way to solve this problem:

There is a way to construct a CW-approximation $Z\to X$ for any path-connected space $X$ such that $Z$ only has a single $0$-cell. This construction can be seen in Hatcher's book in the section on CW-approximations. Then if $X$ is a CW-complex, it follows from Whitehead's Theorem that the approximation is a homotopy equivalence.

  • $\begingroup$ Thanks for the answer. I also thought about using CW-approximation, but my aim is to prove a theorem done in class that was before CW-approximation, so I wanted to avoid it. $\endgroup$
    – Alessandro
    Dec 28, 2021 at 16:57

You must log in to answer this question.

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