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 $\pi_1$?
induced homomorphism on fundamental group
2
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algebraic-topology
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0I 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 $S^1=\{z\in\mathbb C\mid |z|=1\}$.) – 2011-07-07
1 Answers
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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.
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0:thanks alot it's clear now. the only surjective homomorphism $\mathbb Z\rightarrow \mathbb Z$ is multiplication by $1$ and $-1$. Actually $Aut(\mathbb Z)=\mathbb Z_2$. – 2011-07-07