The question has been answered in the comments. To justify writing this answer I'll sweeten it with some links and further details.
Compact manifolds (possibly with boundary) have the homotopy type of a CW-complex, see the answers to this MO-question which also provide links to the literature.
If $f:X\to Y$ induces isomorphisms on cohomology and $X$ and $Y$ are simply-connected spaces, then $f$ is a weak equivalence, discussed e.g. here. Subtleties with homological Whitehead theorems in the non-simply-connected case are discussed in this MSE-question. I suppose that the implication from cohomology isomorphism to weak equivalence needs isomorphisms on fundamental groups and cohomology with all local systems coefficients, unless the space is nilpotent, see this paper of Gersten and the cited paper of E. Dror.
If $f:X\to Y$ is a weak equivalence of CW-complexes, then it is a homotopy equivalence, and therefore has a homotopy inverse $g:Y\to X$, which induces the inverse cohomological isomorphism for $f^\ast:H^\ast(Y)\to H^\ast(X)$. By the first remark this applies to all compact manifolds with boundary.
Finally, an example with homology spheres, some other examples of what can go wrong have been provided in the links above. In Milnor's paper on Brieskorn homology spheres, one can find lots of examples of homology spheres which have a contractible universal cover (they are covered by hyperbolic three-space and their fundamental groups are related to triangle groups). The plus-construction again provides a map $M\to S^3$ from the homology sphere to $S^3$. (The plus-construction produces a CW-complex which by Whitehead's theorem again is homotopy equivalent to $S^3$.) This map $M\to S^3$ induces an isomorphism on cohomology with integral coefficients. However, there can not be an inverse, because the universal covering of $M$ is contractible and so every map $S^3\to M$ has degree $0$. This provides another example that some condition like simply-connected is necessary.