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Suppose $S$ is a closed surface with genus greater than $1$. Give an example of two loops on $S$ which are not homotopic, but are homologous to each other.

I need help solving this question. I have read the Wikipedia page on it and I realize that it does not contain the necessary knowledge to visualize this problem.

And I was also wondering the relation between the fundamental group $\pi_1(S)$ and the first homology group $H_1(S)$? Some restrictions is that $S$ is a closed surface and the genus is greater than $1$.

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As OP seems to require further elaboration on the torus case, here are some details. Consider the standard presentation of the torus:

The tricky thing to realize is that the loop $ABA^{-1}B^{-1}$ going around the square is actually homotopic to the constant loop by scaling the loop down into the top left corner, say. In the same spirit, you can find a homotopy between $AB$ and $BA$ by pulling the loop across the square.

But this only worked because we used all of the $2g = 2$ generators and considered the loop around the full polygon. For example, consider the $g=2$ case that we have in the following image:

We can again scale down the full loop $a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}$ down to one of the points. But there is clearly no homotopy between $a_1b_1$ and $b_1a_1$. You’ll get stuck across the polygon when you try to homotope these two.

The homology group $H_1(S)$ is the abelianization of the fundamental group. So for instance, on the two-holed torus, draw it as an octogon with the usual gluing rules, i.e. the edges go $e_1 e_2 e_1^{-1} e_2^{-1} e_3 e_4 e_3^{-1} e_4^{-1}$.

Then $e_1e_2$ is homologous to $e_2e_1$ but not homotopic.

If $\alpha,\beta$ are loops which are not homotopic as based maps from the circle, then the product (in fundamental group terms) $\alpha\beta^{-1}$ is not homotopic to the constant loop. Let’s suppose that $[\alpha]$ and $[\beta]$ also do not commute (which is the case for generating elements of surface groups with genus greater than $1$).

The relationship between the fundamental group and the first homology group of a (connected) space is that $\pi_1(X)^{ab}=\pi_1(X)/[\pi_1,\pi_1]\cong H_1(X)$ where $[\pi_1,\pi_1]=\langle xyx^{-1}y^{-1}\mid x,y\in\pi_1(X) \rangle^{\pi_1(X)}$. That is, the first homology group is the abeliantisation of the fundamental group.

Given the first sentence then, $\alpha\beta\alpha^{-1}\beta^{-1}$ is not homotopic to the constant loop unless the fundamental group possibly has $2$-torsion which doesn’t occur for orientable surfaces. But note that abelianised we have $$\mbox{ab}([\alpha\beta\alpha^{-1}\beta^{-1}])=\mbox{ab}([\alpha][\beta][\alpha^{-1}][\beta^{-1}])=0$$ and so the loop $\alpha\beta\alpha^{-1}\beta^{-1}$ is homologous to the constant loop.

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