is there a continuous surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H1(T,x) \rightarrow H1(T,x)$is the zero-map.

Question

Let $T = \mathbb{S}^1 \times \mathbb{S}^1$ be a torus and $x \in T$. Prove or disprove: There exists a continuous surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H_1(T,x) \rightarrow H_1(T,x)$is the zero-map.

I have no idea how to solve this kind of problems. All I know is that since the fundamental group of the torus is abelian we may think of $f^*$ as a map of fundamental groups instead. We can also say by the lifting lemma that our maps lifts to its universal cover $\mathbb{R}^2$ which is contractible, but I don't know if this is relevant.

Any help will be much appreciated

I guess its impossible for $S^1$ since it a path that wraps $S^1$ hence its not zero — TheGeometer, May 11 '18 at 04:15
My hunch is you can postcompose with a surjection to the circle to see that map must not be zero. — Max, May 12 '18 at 02:12
@TheGeometer: It's not impossible with $S^1$! First flatten your circle to an interval and then wrap the interval around the circle. So intuitively, your path travels around the circle so it is surjective, but then backtracks back to where it started, so it is degree $0$. — Cheerful Parsnip, May 12 '18 at 20:37

score 5 · Accepted Answer · answered May 12 '18 at 03:03

Here is an outline on constructing such a map.

I'm thinking of $S^1 = \{z\in \mathbb{C}: |z|=1\}$. Then define $g:S^1\rightarrow S^1$ by $g(z) = \begin{cases} z^2 & \operatorname{Im}(z)\geq 0\\ \overline{z}^2 & \operatorname{Im}(z)\leq 0\end{cases}$.

Note that if $\operatorname{Im}(z) = 0$, then $z = \pm 1$, so $z^2 = \overline{z}^2 =1$ in this case. Thus, $g$ really is continuous.

Intuitively, $g$ wraps the top half of $S^1$ fully around $S^1$ one way, and wraps the bottom half of $S^1$ fully around $S^1$ the other way.

Can you prove $g$ is surjective? Can you prove that $g$ is the $0$ map on $H_1$? Can you use $g$ to construct $f:T\rightarrow T$ with the properties you're looking for?

score 3 · Answer 2 · answered May 12 '18 at 11:59

3

You can try also the following solution: Take any surjective continuous map from $T$ to $I$ - unit interval, for example a height function. Then using Peano curve you can find surjective continuous map from $I$ to $I^2$, and then the quotient map from $I^2$ to $T$. This map will be continuous and surjective, and induces $0$ in $H_1(X)$, since it factors through contractible space.

Actually, this map is null-homotopic.

To be totally fair, my friend gave me this solution, so I am not claiming I had this idea myself. Thanks Andrzej :)

answered May 12 '18 at 11:59

Igor Sikora

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This works but nothing this complicated is needed. – Cheerful Parsnip May 12 '18 at 23:34
Maybe not, but that is basically the same idea as yours in the last comment under the post :) – Igor Sikora May 13 '18 at 00:08
Well, mapping to $S^1$ you don't need a space-filling curve. Then to do $S^1\times S^1$, just use my idea in each coordinate. Upvoted your answer, btw. – Cheerful Parsnip May 13 '18 at 00:35
Yes, now I see. I can project a torus onto a square, and then back on a torus. Indeed simpler then using Peano curves – Igor Sikora May 13 '18 at 09:34

is there a continuous surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H1(T,x) \rightarrow H1(T,x)$is the zero-map.

2 Answers2