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I am trying to prove that ($S^1\times\{1\})\cup(\{1\}\times S^1)$ is not a retract of $S^1\times S^1$. Any help would be appreciated.

Thank you!

Martin Citoler
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jmcats
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  • What aspects of topology have you studied thus far? Homotopy theory? Homology? We could proceed by contradiction and suppose that if one is a retract of the other, then some aspect that is preserved under retraction is different for the spaces. In this case, you could look at how one space is connected while the other is not. – Kris Williams May 02 '13 at 21:48
  • Thank you Kris. We have studies Homotopy theory. I will give it a try. – jmcats May 02 '13 at 22:16
  • @Kris: The second space is the torus, which is connected, but the first one seems path connected, hence connected to me. Both sides of the union are circles hence path connected. Moreover, $(1,1)\in S^1\times{1}$ and $(1,1)\in{1}\times S^1$, so a point of $S^1\times{1}$ can be connected with a point from ${1}\times S^1$ by a path via $(1,1)$. – HSN May 02 '13 at 22:26

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If there were a retraction, you would get a composition $$S^1\vee S^1 \subset S^1\times S^1\overset{r}{\to}S^1\vee S^1$$ which is the identity. On the level of $\pi_1$ you would get $$F_2\to \mathbb Z\oplus\mathbb Z\to F_2$$ being the identity homomorphism on the free group on $2$ generators, $F_2$. This is impossible, for example, because all commutators would lie in the kernel of the left-hand map.

Cheerful Parsnip
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  • In case OP doesn't know about wedge sums yet: $S^1\vee S^1$ is the disjoint union of $S^1 with S^1$ where the basis points of both circles are identified. It is therefore "the same" as the $S^1\times{1}\cup{1}\times S^1$ occuring in your question. – HSN May 02 '13 at 22:32