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Given a $2\times2$ matrix $A$ with entries in $\mathbb Z$, it acts on \begin{equation} \mathbb T^2=\mathbb R^2/\mathbb Z^2 \end{equation} and then leads to a map \begin{equation} A_*\colon H_2(\mathbb T^2)\to H_2(\mathbb T^2) \end{equation}

I want to know is this map just the scalar multiplication by $\det A$?

Can someone suggest a neat proof?

Thank you in advance.

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  • Very brief idea: $H_2(\Bbb T^2)$ is induced by the degree map on cycles. – Arthur Nov 25 '18 at 18:07
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    Yes. To my mind it's easier to see the corresponding fact for cohomology $H^2$ (because then you can use the cup product), then apply UCT. – Qiaochu Yuan Nov 25 '18 at 19:04

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