# Isomorphism in all homology and cohomology groups

Let $$M^n$$ be a closed, connected and orientable topological manifold of dimension $$n \geq 2$$ and let $$f : M \to M$$ be a continuous map. Assume that $$f_* : H_n(M) \to H_n(M)$$ is an isomorphism. The exercise is:

Show that $$f_* : H_q(M;G) \to H_q(M;G)$$ and $$f^* : H^q(M;G) \to H^q(M;G)$$ are isomorphisms for every $$q \geq 0$$ and every abelian group $$G$$.

The book suggests to start with the case $$G = \mathbb{Z}$$. I already showed that $$f^* : H^n(M) \to H^n(M)$$ and $$f_* : H_0(M) \to H_0(M)$$ are isomorphisms (using Poincaré duality and the Universal Coefficient Theorem), but I don't know how to proceed. For example, how do we show that $$f_* : H_1(M) \to H_1(M)$$ is an isomorphism?

PS.: I managed to show that $$f^* : H^k(M) \to H^k(M)$$ is injective and $$f_* : H_k(M) \to H_k(M)$$ is surjective for every $$k$$. Probably I am missing an algebraic argument to conclude. Could you help me? My thought was to write $$H^k(M) = B^k \oplus T^k$$, with $$B^k$$ free abelian and $$T^k$$ the torsion subgroup. Since $$f^*$$ is injective, we have $$\operatorname{rank}(H^k(M)) = \operatorname{rank}(\operatorname{im} f^*)$$, but this doesn't necessarily imply that $$f^*$$ is surjective.

• Consider the pairing $H_k\times H_{n-k} \to H_n$. If the map induced by $f$ was non injective on $H_k$, say $fv = 0$, then pick $w \in H_{n-k}$ so that $(v,w)$ generates $H_n$ but then $0 = (fv,fw) = f(v,w)$ so that the induced map on $H_n$ is zero. – Asvin Jun 26 '19 at 2:30
• Actually, using the pairing $H_k \times H^k \to H_n$ will let the previous argument go through in general (the old argument only worked for fields). – Asvin Jun 26 '19 at 2:33
• @Asvin are you sure the indices are correct? What pairing are you referring to? – Eduardo Longa Jun 26 '19 at 2:37
• The Kronecker pairing. But it's really just taking a cocycle, a cycle and evaluating the cocycle on the cycle. I am not totally sure that it lands in $H_n$ which is why these are comments. – Asvin Jun 26 '19 at 2:47
• It lands on $\mathbb{Z}$ – Eduardo Longa Jun 26 '19 at 2:52

Here's the main idea of the proof. By Poincaré duality, given a nonzero class $$\alpha\in H^q(M)$$, you can expect there to be a class $$\beta\in H^{n-q}(M)$$ such that $$\alpha\beta\in H^n(M)$$ is nonzero. So, since $$f^*$$ is an isomorphism on $$H^n(M)$$, $$f^*(\alpha\beta)=f^*(\alpha)f^*(\beta)$$ is nonzero so $$f^*(\alpha)$$ is nonzero. Thus $$f^*$$ is injective on $$H^q(M)$$ and you can hope to then formally deduce that it must also be surjective using finiteness properties (for instance, an injective endomorphism of a finite dimensional vector space is automatically surjective).
Now there are complications to actually carrying this out: for instance, such a $$\beta$$ need not actually exist if you're taking cohomology with coefficients in $$\mathbb{Z}$$ (it won't if $$\alpha$$ is a torsion class). There are various ways to handle this, but I think the easiest is to actually not start with coefficients in $$\mathbb{Z}$$ as your book suggests but instead work with coefficients in a field. Then the statement of Poincaré duality is nicer (you really do just have a perfect pairing on finite-dimensional vector spaces given by the cup product, and also cohomology is simply the dual vector space of homology). Then you can deduce the case of $$\mathbb{Z}$$ coefficients from the case of all fields and deduce the case of arbitrary coefficients from there.
• There are a few ways you can do that. For instance, you can think about what the homology of the mapping cone of $f$ could possibly be, using the universal coefficient theorem and the classification of finitely generated abelian groups. – Eric Wofsey Jun 26 '19 at 21:04