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Given a topological space $X$, I can show that the following are equivalent:

  • X is contractible (that is, has the homotopy type of a point)
  • There is some $x \in X$ such that $\{x\}$ is a deformation retract of $X$ (where the homotopy between the retraction $X \to \{x\}$ and the identity on $X$ needn't be fix $x$)

In Steven Weintraub's 'Fundamentals of Algebraic Topology', on pg. 4 is written 'A space is contractible if it has the homotopy type of a point *, or, equivalently, if for some, and hence for any, point $x_0 \in X$, $x_0$ is a deformation retract of $X$.'

So I'm trying then to show that the above two conditions imply that any point of $X$ is a deformation retract of $X$, but I'm unable to do this.

Say $x_0 \in X$ is a deformation retract of $X$, so that $if$ where $f$ is the constant map $f: X \to \{x_0\}$ and $i$ is the inclusion $\{x_0\} \hookrightarrow X$ is homotopic to the identity on $X$. I don't see how from this information I can get a homotopy between the identity on $X$ and $ig$ where $g$ is the constant map $g: X \to \{x_1\}$ for some $x_1 \in X$.

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If $X$ is contractible, then there is a map $i:\{*\}\to X$ and a map $f:X\to\{*\}$ (where $\{*\}$ is the one point space) such that $if\simeq\mathbf 1_X$. If $x_1$ is any point of $X$, we can set $j:\{*\}\to X, j(*)=x_1$, and then, because we can concatenate homotopies and compose them with maps, $$jf\simeq (if)jf=i(fj)f=if\simeq\mathbf 1_X$$

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  • 1
    $\begingroup$ I see, I should've worked with contractibility, thanks. $\endgroup$ – user212931 Feb 4 '15 at 22:38

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