I try to solve the problem: Let's we have a map from torus to torus $f: \mathbb T \to \mathbb T$ that inuducing homomorphism between first groups of homology: $f_*: H_1(\mathbb T) \to H_1(\mathbb T)$ wich has a $2\times2$ matrix $A$. I need to prove that the induced homomorphism between second homology groups is a multiplicaton by $det(A)$. I have biult one of such a map that just transfers both of generators of the $H_1(\mathbb T)$ to the linear combination of the two generators of the $H_1(\mathbb T)$ of the other torus. Such a map is obviuosly inducing $f_*: H_1(\mathbb T) \to H_1(\mathbb T)$ wich has a $2\times2$ matrix $A$ and $f_*: H_2(\mathbb T) \to H_2(\mathbb T)$ that is multiplication by $det(A)$. Then, my idea is to show that the second homology homomorphism in such situation depends only on the first homology homomorphism (I tried use Mayer-Vietoris for this and degree theory but failed). Can you please help with it?
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Do you know Poincare duality? – Moishe Kohan Nov 28 '19 at 14:12
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$H_2(\mathbb T)\simeq H^0(\mathbb T)$? But what does it give us? – Igor Kozyrev Nov 28 '19 at 16:14
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1No, I meant the duality $H^1\cong H_1$, or the intersection form on $H_1$, or the cup product $H^1\otimes H^1\to H^2$. – Moishe Kohan Nov 28 '19 at 17:32