null homotopy of 1-skeleton of simply connected simplicial complex induced by homotopy identity

Question

I am reading a book suggesting that the following is true:

Let $X$ be a simply connected (maybe finite) simplicial complex. Then there exists a continous map $$g : X \to X,$$ homotopic to the identity, which maps the 1-skeleton of $X$ to a point.

A confirmation or counterexample would be great.

Answer 1

There is this corollary in Hatcher's book Algebraic Topology:

Corollary 4.16. If $(X,A)$ is an $n$-connected CW-pair, then there exists a CW pair $(Z,A) \simeq (X,A)$ such that all cells of $Z - A$ have dimension greater than $n$.

This is proven using cellular approximation techniques. Now if you choose an arbitrary base point $x \in X$, the pair $(X,x)$ is $1$-connected because $X$ is simply connected. Thus there exists a CW pair $(Z,x) \simeq (X,x)$ such that all the cells of $Z$ (other than $x$) have dimension at least $2$. Let $f : Z \to X$ be the homotopy equivalence. By Whitehead's theorem, this is a strong homotopy equivalence, so let $g : X \to Z$ be the homotopy inverse.

Now let $\varphi : X \to X$ be the composition $fg$, so that $$\varphi \simeq \operatorname{id}.$$

By using the cellular approximation theorem (Theorem 4.8 in Hatcher), we can assume that both $f$ and $g$ are cellular. In particular, the whole $1$-skeleton of $X$ gets mapped by $g$ to the $1$-skeleton of $Z$, which is just $\{x\}$, a point. Therefore $\varphi(X^{(1)}) = \{ f(x) \}$ is also just a point.