Let $f:N\_1\to N\_2$ be a homotopy equivalence between two simply connected manifolds with boundary of the same dimension. Can it be extended to a homotopy equivalence between closed manifolds $f:M\_1\to M\_2$ such that $N\_i\subset M\_i$? (I suppose there are counterexamples, but what I am actually interested in are the obstacles.)



Extending homotopy equivalence between manifolds with boundary
The answer in general is no. Let $N\_1$ be $S^2\times D^2$, and $N\_2$ be the disk bundle of Euler class $1$ over $S^2$. Then both are homotopy equivalent to $S^2$. But no homotopy equivalence will extend to closed manifolds. For if $f: M\_1 \to M\_2$ is a homotopy equivalence, with $N\_i \subset M\_i$, then $N\_i$ carry homology classes $a\_i \in H\_2(M\_i)$ with $f\_\*(a\_1) = a\_2$. But $a\_1 \cdot a\_1 =0$ and $a\_2 \cdot a\_2 =1$ so there can't be such a homotopy equivalence, by homotopy invariance of the intersection form. (This would also work for non-orientable extensions, using the mod $2$ intersection form.)


I don't know a complete answer to your question of `what are the obstacles', but I guess you could say that the intersection form is one such obstacle. The point is that the intersection form is not a homotopy invariant for manifolds with boundary, but it is on closed manifolds, since it's Poincare dual to cup product, which is homotopy invariant. If your homotopy equivalence maps $\partial N\_1$ to $\partial N\_2$ and is also a homotopy equivalence of pairs, then you could just double the $N\_i$ to get a homotopy equivalence. So the best general principle might be: ask how the given $f$ interacts with the boundary.


