Extend a Riemannian metric defined on the boundary. If $M$ is a (compact) Riemannian manifold with nonempty boundary and I have a Riemannian metric defined on $\partial M$, Is it possible to obtain a Riemannian metric on the whole $M$ extending the one on the boundary?
I think that it is possible to extend it to a tubular neighborhood of the boundary by taking it to be diffeomorphic to $\partial M \times [0,1)$. What about the rest of the manifold?
 A: There is a bundle $\mathcal R \to M$, whose fiber over $p$ is the space of inner products on $T_pM$. This is a convex space. Smooth Riemannian metrics are precisely the smooth sections of this bundle.
First, here is a straightforward and less homotopy-theoretic way of doing this. Pick an open cover $U_\alpha$ of your manifold such that $U_0 = \partial M \times [0,1)$ and the other $U_\alpha$ are charts, and let $\rho_\alpha$ be a subordinate partition of unity. Pick a metric $g_\alpha$ on each $U_\alpha$; on $U_0$ let $g_0$ be the metric you constructed, and on the other ones just pull it back from $\Bbb R^n$. Then let $g = \sum \rho_\alpha g_\alpha$. Because convex combinations of inner products are inner products, $g$ is, pointwise, an inner product, and varies smoothly as desired. (This is just a mild modification of the proof that any manifold supports a Riemannian metric.)
But what you're asking is really an obstruction theory problem: given a section of $\mathcal R$ on the subspace $\partial M$, can you extend this section to all of $M$? Here's an obstruction theory answer: put a cell structure on $M$ so that $\partial M$ is a subcomplex. Now, cell by cell, extend the section: given a cell $e_i$ such that the section is defined on $\partial e_i$, trivialize the bundle $\mathcal R$ over $e_i$. (This is possible because $e_i$ is contractible.) Because the space of inner products on a vector space is convex, it is contractible, so the map $\partial e_i \to \text{InnerProducts}$ is null-homotopic; that is, you can extend it to all of $e_i$. Proceeding in this manner you have defined a section over all of $M$. This idea generalizes. Given any bundle $\mathcal R \to X$ with contractible fibers, and a section of the bundle on a subcomplex $Y \subset X$, you can extend the section to all of $X$.
