Invertibility of $-\Delta +1$ on a closed manifold On page 468 of Hamilton's Ricci Flow by Chow et al., the authors claim:
"The operator $-\Delta+1$ is invertible and its inverse $(-\Delta+1)^{-1}:L^2(M)\to W^{2,2}(M)\hookrightarrow L^2(M)$ is a compact operator by elliptic regularity (see for example Gilbarg and Trudinger)." Here, $M^n$ is a closed manifold. 
Of course GT treats only PDE on $\Bbb R^n$, so there's some work to be done here. One solution would be to apply elliptic regularity in charts, as $-\Delta +1$ is an elliptic operator in a chart. But for this we need a boundary condition. 
As far as I understand, the existence of $(-\Delta+1)^{-1}$ means that for any $f\in L^2(M)$, there is a $u\in W^{2,2}(M)$ with $(-\Delta+1)u=f.$ I think the result from GT that I want is 8.9, which says that under mild conditions on the coefficients, the Dirichlet problem can always be solved with $W^{1,2}\cap W^{2,2}_\text{loc}$ regularity of the solution. So if I apply this chartwise, then it would seem the global solution (which we would have to patch together somehow) would actually depend on the choice of charts. So the inverse wouldn't be unique. 
Now, the other way would be to solve $(-\Delta+1)u=f$ weakly on $M$, in the global sense. Then reduce the weak equation on $M$ to a weak equation on charts, and then apply Corollary 8.11 from GT, which gives smoothness of weak solutions. 
What's the correct approach here?
 A: The first approach doesn't quite work because you can't tell in advance which boundary conditions to put locally on $u$. At the end, you'll need some sort of a global argument. Here is one way to do it:
In what follows, I'll assume that $M$ is closed.


*

*Develop first the theory of Sobolev spaces on manifolds. This can be done in variety of ways (more coordinate free or less coordinate free) but because $M$ is compact, all will turn out to be equivalent to each other. The advantage of working with manifolds without boundary is that all the subtle regularity results one needs to impose on the boundary of sets in order for the various embeddings between Sobolev spaces, Holder spaces, etc to hold become irrelevant. You never approach the boundary because there is no boundary so you only need to work with the local embeddings which hold under practically no assumptions.

*Define when a partial differential operator $D \colon C^{\infty}(M) \rightarrow C^{\infty}(M)$ (or more generally an operator between sections of vector bundles over $M$) is elliptic and show the basic a priori estimate. If $D$ is of order $k$, the estimate will look like this:
$$ \| u \|_{H^{n + k}(M)} \leq C \left( \| Du \|_{H^n(M)} + \| u \|_{H^n(M)} \right). $$
This can be done by using a partition of unity and transferring the problem to $\mathbb{R}^m$. Again, the fact that $M$ has no boundary means that one can use the interior elliptic regularity estimates and one doesn't need to care about regularity up to the boundary (because globally there won't be any boundary). At this point we have elliptic regularity for weak solutions.


Given the results above, to prove that $-\Delta + I$ is invertible, we can proceed in one of two ways:


*Using the basic estimate, show that all elliptic operator are Fredholm. Then $\Delta$ is self-adjoint and so $-\Delta + I \colon H^2(M) \rightarrow L^2(M)$ will be a Fredholm operator of index zero. Such and operator is one-to-one iff it is onto so it is enough to show that $-\Delta + I$ is one-to-one which is easy because $\Delta$ is semi-negative-definite. 

*Use the Lax-Milgram lemma globally to deduce existence and uniqueness of $H^1$ solutions to $-\Delta(u) + u = f$ and then upgrade them using elliptic regularity. Here there won't be any boundary conditions involved (the space we will work on will be $H^1(M) = H^1_0(M)$).
