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.



$\pi_1(S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z}$, 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 $S^1 \vee S^1$.

  • $\begingroup$ Do you know a way to prove this only using the fact that the wedge sum is a deformation retract of the punctured torus in order to get a contradiction? $\endgroup$ – user42912 Nov 14 '12 at 4:48
  • $\begingroup$ Are you asking if there's a general fact that $X \setminus \{p\}$ is never a retract of $X$? This is true provided $p$ has a neighborhood homeomorphic to $D^n$ for some $n$, but you'd need to use homology theory -- the computations for $\pi_1$ of a wedge of spaces and for a product of spaces are very basic by comparison. $\endgroup$ – Thomas Belulovich Nov 14 '12 at 12:54
  • $\begingroup$ @BenjaLim see again :) $\endgroup$ – user42912 Nov 15 '12 at 2:14
  • $\begingroup$ can you please add the proof that the induced inclusion map is not injective? Thank you for you answer :) $\endgroup$ – user42912 Nov 25 '12 at 12:14
  • $\begingroup$ just added that above; the map $G \to G^{ab}$ is non-injective whenever $G$ was not itself abelian. $\endgroup$ – Thomas Belulovich Nov 25 '12 at 14:57

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