Does removing a closed contractible subset from a closed connected topological manifold leave the resulting manifold connected? From looking at small examples, it appears that if you remove a closed contractible subset of a closed, connected topological manifold the resulting manifold will be connected. Is this statement true in this level of generality, or does it require additional assumptions (orientability of the manifold, that the manifold is triangularizable, that the closed subset have nonempty interior, that the boundary of the closed subset is homeomorphic to a sphere...)?
Especially nice would be a reference to an existing proof.
 A: This is true and is a consequence of a form of Poincaré duality.  Specifically, if $M$ is a closed oriented $n$-manifold and $K\subseteq M$ is closed, there is an isomorphism $\check{H}^k(K)\cong H_{n-k}(M,M\setminus K)$, where $\check{H}$ denotes Čech cohomology.  Using mod 2 coefficients, the orientability assumption can be dropped.  This in particular implies that if $\check{H}^{n-1}(K;\mathbb{Z}/(2))$ is trivial then so is $H_1(M,M\setminus K;\mathbb{Z}/(2))$, and hence the map $H_0(M\setminus K)\to H_0(M)$ is injective so $M\setminus K$ has at most one connected component contained in each connected component of $M$.  As long as $n\neq 1$, this hypothesis is satisfied when $K$ is contractible, and the case $n=1$ can be verified separately (since then $M$ can only be a disjoint union of circles and $K$ can only be a point or a closed interval on one of the circles).
I don't know a reference for exactly this duality statement, but a similar statement using singular cohomology with coefficients in $\mathbb{Z}$ (and requiring $K$ to additionally be locally contractible) is Theorem 3.44 in Hatcher's Algebraic Topology.  Hatcher's proof can easily be modified to obtain the result for Čech cohomology (since then the cohomology of $K$ can be computed as the direct limit over open neighborhoods of $K$, without needing any special conditions on $K$) and with arbitrary coefficients (as long as $M$ is orientable with respect to those coefficients).
