TITLE: Show that something is not a retract
QUESTION [5 upvotes]: 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!

REPLY [4 votes]: 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.