# Deformation retracts?

• Oct 23rd 2010, 08:51 AM
math8
Deformation retracts?
Let $\displaystyle x_0 \in S^1$.

Is $\displaystyle S^1$ X $\displaystyle \{ x_0 \}$ a retract of $\displaystyle S^1$ X $\displaystyle S^1$?

Is it a deformation retract of $\displaystyle S^1$ X $\displaystyle S^1$?

I would think one would use the fact that if $\displaystyle S^1$ X $\displaystyle \{ x_0 \}$ were a deformation retract of $\displaystyle S^1$ X $\displaystyle S^1$, then their fundamental groups would be isomorphic.

But the fundamental group of $\displaystyle S^1$ X $\displaystyle S^1$ is ZXZ, while I think the fundamental group of $\displaystyle S^1$ X $\displaystyle \{ x_0 \}$ is ZX{0}. Would this be enough to justify it's not a deformation retract? If yes, I have a feeling that $\displaystyle S^1$ X $\displaystyle \{ x_0 \}$ is a retract of $\displaystyle S^1$ X $\displaystyle S^1$. But how do you prove this?
• Oct 23rd 2010, 01:36 PM
HappyJoe
You are quite right.

A deformation retract is a special case of a homotopy equivalence. Hence, if a subspace $\displaystyle A$ was a deformation retract of the space $\displaystyle X$, the two spaces would indeed have the same fundamental group (which they don't in your case).

To show that $\displaystyle S^1\times\{x_0\}$ is however a retract of $\displaystyle S^1\times S^1$, you just use the first map that comes to mind, i.e. use the map $\displaystyle f\colon S^1\times S^1\rightarrow S^1\times\{x_0\}$ given by $\displaystyle f(x,y) = (x,x_0)$, so you map all of the second copy of $\displaystyle S^1$ to the point $\displaystyle x_0$
• Oct 25th 2010, 07:35 AM
math8
Thanks, this makes a lot of sense.