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Thread: Deformation retracts?

  1. #1
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    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?
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  2. #2
    Member HappyJoe's Avatar
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    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 $\displaystyle x_0$
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  3. #3
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    Thanks, this makes a lot of sense.
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