If every finite subcomplex is nullhomotopic, is a simplicial complex contractible?

Question

Suppose $X$ is a CW-complex such that for every finite subcomplex $Y$, $Y$ can be contracted in $X$, that is, the inclusion map $ i \colon Y \to X$ is nullhomotopic. Is it then true that $X$ is contractible?

Answer 1

By Whitehead's theorem, and any map $j \colon * \to X$, where $*$ denotes the one-point space, it is enough to show that $X$ is weakly contractible.

Now, any map $S^n \to X$ has compact image which then must be contained in some finite subcomplex $Y$. Thus, by the given condition, $\pi_n(X)$ is trivial for every $n$.

Obviously the converse is also true. That is, if $X$ is itself contractible, then every $Y$ can be contracted in $X$. However, it need not be true that every $Y$ itself is contractible, for example $S^n \subset S^\infty$. See also, this MO question.