Question: If $X$ is a simplicial complex that's simply connected and $2$-dimensional, does there always exist a contractible subcomplex $Y$ satisfying $X^{(1)} \subseteq Y$?

The statement is true "down a dimension": If $X$ is connected and $1$-dimensional, then there exists a contractible subcomplex $Y$ satisfying $X^{(0)} \subseteq Y$, namely you can use a spanning tree. So, one could think of the desired $Y$ in the original problem as a $2$-dimensional analog of a spanning tree, with contractibility being the key desired property.

It seems there is a higher dimensional analog of "spanning tree" in the literature, at least for finite complexes, e.g., Definition 3.1 of this. Note that Proposition 3.7 of that paper implies that (finite) simply connected $2$-complexes indeed have these sorts of $2$-dimensional "spanning trees". But these are not necessarily contractible, they're more of a homological analog of trees.

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