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Show that the map $\phi_k : z \mapsto z^k$ from $S^1$ to $S^1$ induces multiplication by $k$ on $H_1.$

This question is appeared in the midterm exam at our institute which I am unable figure out properly as how to proceed. Any help would be highly appreciated.

Thanks!

Fanatics
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1 Answers1

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Well, the answer to this question can be quite short - or quite long depending on your familiarity with algebraic topology. The short answer is: View the generator of $H_1(S^1)$, call it $\gamma$ as the loop which goes around $S^1$ precisely once. Then, the morphism $z \mapsto z^k$ will make $\gamma$ wrap around your $S^1$ precisely $k$-times, and therefore maps $\gamma$ of $H_1(S^1)$ to $k \gamma$. Since it induces multiplication by $k$ on the generator of $H_{1}(S^1)$ it must induce multiplication by $k$ on the whole of $H_1$.

This also agrees with the idea that the degree of a map is equal to the number of pre-images of a generic point - look at the map $z \mapsto z^{k}$, and choose a random point, say $p$, on $S^{1}$ in the image. Then, there are $k$ points in the pre-image which get mapped to $p$ which are $\frac{2\pi}{k}$ radians apart.

Locally unskillful
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  • How do you know that if a morphism wraps around the circle $j$-times then it will induce multiplication by $k$ on the generator of $H_1(S^1).$ Aren't you using any specific relation between $\pi_1(S^1)$ and $H_1(S^1)$ here? – Fanatics Jun 30 '21 at 15:16