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I'm trying to prove or disprove that the wedge sum of two circles is a retract of the torus. Intuitively it seems true, because the torus is defined as $S^1\times S^1$. I tried to disprove also, but the only tools I know to do this is if one space was simply connected and the another don't, which is not the case.

Thanks

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$\pi_1(S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z}$, the free product. There's a generating loop for each copy of $S^1$, but these don't commute.

On the other hand, $\pi_1(S^1 \times S^1) = \mathbb{Z}^2$. The generating loops on the torus can be passed over each other and so this group is abelian.

The inclusion map $\pi_1(S^1 \vee S^1) \to \pi_1(S^1 \times S^1)$ is indeed the abelianization. This is not injective (the image of $\alpha\beta\alpha^{-1}\beta^{-1}$ is zero), so $S^1 \vee S^1$ is not a retract of the torus.

However, it is the case that $S^1 \vee S^1$ is homotopy equivalent to a punctured torus; in fact you can arrive at $S^1 \times S^1$ by attaching a single 2-dimensional cell to $S^1 \vee S^1$ via an attaching map corresponding to the loop $\alpha \beta \alpha^{-1} \beta^{-1}$, the commutator of the two generating loops of $S^1 \vee S^1$.

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    just added that above; the map $G \to G^{ab}$ is non-injective whenever $G$ was not itself abelian.2012-11-25