TITLE: Inverse cohomological isomorphisms
QUESTION [8 upvotes]: Let $\ M'\ M''\ $ be simply-connected Hausdorff compact manifolds (possibly with boundary for another variant of the question). Let $\ f:M'\rightarrow M''\ $ be a continuous function which induces an isomorphism of the cohomological rings. Does there exist a continuous function $\ g:M''\rightarrow M'\ $ which induces the inverse cohomology ring isomorphism?

REPLY [7 votes]: Some further comments on homology spheres:  First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point.  In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$.  If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_n(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere.  (These are classical arguments, incidentally.)