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Let $X$ be a topological manifold with boundary. What is the idea behind the fact that removing the boundary doesn't change the homotopy type of the manifold; i.e.,that is the manifold $X$ has the same homotopy type of $X-\partial X$?

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In a smooth manifold, there is a neighborhood of the boundary $V$ which is diffeomorphic to $\partial M\times [0,1)$, called a collar neighborhood. (Google "Collar neighborhood theorem.") Removing the boundary would give you $\partial M\times (0,1)$. These are homotopy-equivalent in a way that fixes $\partial M\times [.5,1)$, so the homotopy equivalence extends to the interior of the manifold $M$. This homotopy equivalence can be visualized by collapsing both $\partial M\times (0,1)$ and $\partial M\times [0,1)$ down to $\partial M\times [.5,1)$.

For a topological (non-smoothable) manifold, I believe the collar neighborhood theorem also holds, but I'm not as sure about that.

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    You define all of your maps to be the identity away from the boundary of the manifold. That way, using the gluing lemma, you can glue together the maps relating $M$ and $M\setminus\partial M$ and the homotopies between their compositions.2012-10-13