Let $S^1$ be the circle. Let $X= (S^1\times S^1)/(x,y)\sim (y,x)$. An element in $X$ is denoted $[x,y]$. Why the diagonal map $S^1\rightarrow X; x\mapsto [x,x]$ can not be surjective on $\pi_1$?
Asked
Active
Viewed 804 times
2
-
what is "surjective on $\pi_1$"? – Peter Patzt Jul 07 '11 at 20:01
-
@Peter Patzt : the induced homomorphism on fundamental group is not surjective. – palio Jul 07 '11 at 20:24
-
@Palio: I hope this is not too trivial of a question (I have been absent-minded , so I prefer to ask), but I guess you mean that the map induced by $\pi_1$ is not surjective, right? – gary Jul 07 '11 at 21:07
-
This can often be proved by looking at the abelianization of the $\pi_1$'s (and the induced map there), and this is just $H_1$. What can you say about that? (Use cellular homology.) – Aaron Mazel-Gee Jul 07 '11 at 21:08
-
I think $\tau(t)=[e^{2\pi i t},e^{-2\pi it}]$ should not be in the image of the diagoal map, but i am unsure how to prove that. (I considered $S^1={z\in\mathbb C\mid |z|=1}$.) – Peter Patzt Jul 07 '11 at 21:18
1 Answers
2
Your space $X$ is the Moebius band, and the diagonal map is a homeomorphism onto the boundary of $X$. The class of the boundary in $\pi_1(X)\cong\mathbb{Z}$ ($X$ is homotopy equivalent to $S^1$) is $2$, as it goes twice around.
user8268
- 21,348
-
:thanks alot it's clear now. the only surjective homomorphism $\mathbb Z\rightarrow \mathbb Z$ is multiplication by $1$ and $-1$. Actually $Aut(\mathbb Z)=\mathbb Z_2$. – palio Jul 07 '11 at 22:20