I want to show that every map $S^2$ $→$ $T$ induces the zero map on $H_2$. Here $T$ denotes the torus. I know that $H_2(S^2)= H_2(T)=\mathbb{Z}$ but other than that I am not sure how to approach this.
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See the answer here – May 17 '17 at 09:44
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2How about using the long exact sequence of homotopy groups associated to a fibration? Or in a more elementary way the lifting property for maps with value in the base space of a covering. – Ahr May 17 '17 at 09:53
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@A.Rod That's an answer, and it shows more, namely that every map $S^2 \mapsto T$ is homotopic to a constant. – Lee Mosher May 17 '17 at 10:47
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Shall I convert the comment into an answer? I mean there's not much more to say. – Ahr May 17 '17 at 12:14
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@A.Rod: Yes, you should write an answer along these lines. – Michael Albanese May 17 '17 at 17:40
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The point is that you have a covering map $p:\mathbb{R}^2\to T$ whose fiber is $\mathbb{Z}^2$ which is of course discrete. Now if you take a map $f:\mathbb{S}^2\to T$ because our spaces are locally as nice as we want and because $\mathbb{S}^2$ is 1-connected you get a lifting of $f$ to $\mathbb{R}^2$ (prescribed by choosing $t_0\in T$ and $s_0\in f^{-1}(t_0)$ and $x_0\in p^{-1}(t_0)$) that makes the following diagram commutes $$\begin{matrix} & & \mathbb{R}^2\\ &\nearrow &\downarrow \\ \mathbb{S}^2 & \to & T\end{matrix}$$ This means that $H_2(\mathbb{S}^2,\mathbb{Z})\to H_2(T,\mathbb{Z})$ factors through $H_2(\mathbb{R}^2,\mathbb{Z})=0$. Thus the map is $0$.
Incidently here's another way to see it, the fact that you have a fibration $p:\mathbb{R}^2\to T$ gives you a long exact sequence of homotopy groups (where I'm omitting base points)$$\pi_2(\mathbb{Z}^2)\to \pi_2(\mathbb{R}^2)\to \pi_2(T)\to \pi_1(\mathbb{Z}^2)\to \pi_1(\mathbb{R}^2)$$ this tells you that $\pi_2(T)=0$ and thus every map from $\mathbb{S}^2 \to T$ is null-homotopic, in particular it will induce the zero map on all homology groups.
Ahr
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