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Let $X$ be a connected space and $f:X\longrightarrow X$ a map. Suppose $\pi_1(X)$ is an abelian group and that $\pi_1(g):\pi_1(X)\longrightarrow\pi_1(X)$ is an isomorphism. I know we can deduce that $H_1(g):H_1(X)\longrightarrow H_1(X)$ is an isomorphism.

  • Can we deduce that $H_1(g,\mathbb{Z}_2):H_1(X,\mathbb{Z}_2)\longrightarrow H_1(X,\mathbb{Z}_2)$ is an isomorphism?
  • Can we deduce that $H^1(g,\mathbb{Z}_2):H^1(X,\mathbb{Z}_2)\longrightarrow H^1(X,\mathbb{Z}_2)$ is an isomorphism?
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From Universal Coefficient Theorem we have the following natural s.e.s. for homology we have $$ 0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \to H_1(X;\mathbb{Z}_2) \to \text{Tor}(H_0(X;\mathbb{Z});\mathbb{Z}_2)\to 0$$ since $H_0(X;\mathbb{Z})$ is $\mathbb{Z}$ the s.e.s becomes $$ 0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \cong H_1(X;\mathbb{Z}_2) \to 0 $$ and so if a map induces an iso on $H_1(X;\mathbb{Z})$ (and therefore even after $-\otimes\mathbb{Z}_2$) then it has to induce an isomorphism even on $\mathbb{Z}_2$ coefficients (write down the appropriate commutative square and it will be clear).

One cal also avoid the use of UCT for homology making use of the last part of the l.e.s. associated to the s.e.s $$0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}_2\to 0$$ (bockstein) + naturality + $5$ Lemma.

Using UCT for cohomology one has the claim for the other case (left as an exercise :) )

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  • $\begingroup$ Is there any way to do it without UCT for homology? Also, using UCT for cohomology I have that $H^1$ is isomorphic to $\text{Hom}(H_1,\mathbb{Z}_2)$, but I can't conclude. $\endgroup$ – user337924 May 14 '16 at 14:33
  • $\begingroup$ @user337924 I updated the answer. for the second doubt, well you are almost done, just use again naturality :) (or the fact $\hom( - ; \mathbb{Z}_2)$ is a functor and so it sends iso to iso) $\endgroup$ – Riccardo May 14 '16 at 15:16
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The important thing is that all $H_1(-;V)$, and $H^1(-;V)$ are functorial in the group. That is, if you have a group homormophism $G\to H$, then it induces $H_1(G;V)\to H_1(H;V)$ respecting composition, inversions and identities. In particular it sends isomorphisms (of groups) to isomorphisms (of (co)homology).

In particular, when given a $X\to Y$ which induces a fundamental group isomorphism, it induces one on $1$-(co)homology with arbritrary coefficients.

Note that I used group cohomology here, but it is also easy to see that by universal coefficients and Hurewicz, you only use combinations of funtors, such as abelianization, tensoring, and hom.

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