Let $i:S^1 \vee S^1 \rightarrow S^1 \times S^1$ be the inclusion of the figure eight in the torus. We can consider the following induced homomorphisms:

$i_\star :\pi_1 (S^1 \vee S^1) \rightarrow \pi_1 (S^1 \times S^1)$

$j_\star: H_1(S^1 \vee S^1) \rightarrow H_1 (S^1 \times S^1)$

Which, if we calculate them using Van Kampen's Theorem these are just:

$i_\star :{\text{free group on two generators}} \rightarrow \mathbb{Z} \times \mathbb{Z} $

and since we know $H_1$ is the abelianization of $\pi_1$ we also have

$j_\star: {\text{free abelian group on two generators}} \rightarrow \mathbb{Z} \times \mathbb{Z} $

Since $\mathbb{Z} \times \mathbb{Z}$ is already abelian.

Now, my question is, what more can I say about the homomorphisms $i_\star$ and $j_\star$ ? All I know is that $i_\star$ is not injective, since if $a$ and $b$ are the generators of the free group, then $i_\star (ab)=i_\star (a) i_\star (b)=i_\star (b) i_\star (a)=i_\star (ba)$ where the third equality follows from the fact that $\mathbb{Z} \times \mathbb{Z}$ is abelian, but $ab\neq ba$, so the map cannot be injective. I do not know anything more. Is there a way to explicitly describe the induced homomorphisms? If so is this in genera possible? Any help is appreciated. Thank you.

  • $\begingroup$ How is the figure eight mapped into the torus? $\endgroup$ – Steve D May 20 '17 at 0:00

I am assuming the inclusion is along the usual CW decomposition of the torus: a single $0$-cell, two $1$-cells, and a single $2$-cell. By the van Kampen theorem, one may prove that the fundamental group of a space can be obtained from the fundamental group of the $1$-skeleton, modulo the relations given by the attaching maps of the $2$-cells (this is in chapter 1 of Hatcher). So, $i_*$ is actually a surjection onto $\pi_1(S^1\times S^1)$, taking the two generators of $\pi_1(S^1\vee S^1)$ to independent generators of $\mathbb{Z}\times\mathbb{Z}$. If $a,b$ are generators, then the kernel is the normal subgroup generated by $aba^{-1}b^{-1}$.

The abelianization is functorial and right exact, so $j_*$ must be surjective, too. Injectivity follows from the fact that the kernel of $\mathbb{Z}^2\to \mathbb{Z}^2$ is free and therefore $0$, otherwise $\mathbb{Z}^2$ would have rank less than $2$ or have torsion.

One can also descend to the level of cycles and see that each $1$-cycle in $S^1\times S^1$ is a sum of cycles included by $i$, so $j_*$ is surjective. I'm not sure if there is a better argument for injectivity than the one above (the only surjective $\mathbb{Z}^2\to\mathbb{Z}^2$ is injective).


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