Is the boundary of a $k$-simplex missing some $(k-1)$-faces contractible?

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An $k$-simplex $T$ is the convex hull of $k+1$ affinely independent vectors in $\mathbb{R}^k$ (for example, $0$ and the $k$ standard basis). And any convex hull formed by $k$ of them, which itself is a $(k-1)$-simplex, is called a $(k-1)$-face.

Let $S$ be the complex formed by removing the interior of $T$ (so that $S$ is homeomorphic to $S^{k-1}$.) After removing one $(k-1)$-face from $S$, the resulting space is contractible.

I am wondering after removing some $(k-1)$-faces from $S$, is it still contractible? If not, for the resulting space, can the minimum $j$ such that its $j$th reduced Betti number (over rational number field) be strictly smaller than that of $S^{k-1}$ (which is $k-1$)?