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In Hatcher's book on Algebraic Topology in Chapter 0 (where the Homotopy Extension Property is discussed) the claim is being made that there is no retraction of $I^2$ into $I \times \{0\} \cup A \times I$ where $A = \{0, 1, \frac 12 , \frac 13, \ldots \}$. However I couldn't come up with a proof. One can define a retraction if there is only a finite number of "teeth in the comb". It's not clear to me why continuity fails (presumably) at 0 if one tries to extend this retraction to the infinite case. I hope someone can help out.

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2 Answers 2

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If $f$ is a retraction, then it must map $(0,1)$ to $(0,1)$. Since it is continuous, there must a ball $A$ centered on $(0,1)$ such that $f(A)$ is contained in the ball $B$ of radius $\frac 12$ centered on $(0,1)$.

However $A$ must contain a point of the form $(\frac 1n,1)$ for some $n$ (which also maps to itself under $f$), and then the points $(\frac tn,1)$ for $0\le t\le 1$ all lie in $A$ too. Then $t\to f(\frac tn,1)$ is a path in the comb from $(0,1)$ to $(\frac 1n,1)$. Since there is no such path in the comb that doesn't wander outside $B$, $f$ cannot exist.

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Let $C$ be the comb space, and suppose that $r:I^2\to C$ is a retraction. Let $\pi:I^2\to I$ be the projection to the first coordinate, and define $f:I\to I:x\mapsto\pi\big(f(\langle x,1\rangle)\big)$; clearly $f$ is continuous, and $f(x)=x$ for $x\in\{0\}\cup\{1/n:n\in\Bbb Z^+\}$.

For $n\in\Bbb Z^+$ let $p_n=\left\langle\frac1n,1\right\rangle$. Let $B_n$ be an open ball centred at $p_n$ of radius less than $\frac1n-\frac1{n+1}$; since $r$ is continuous, there is an open ball $B_n'$ of some radius $\epsilon_n$ centred at $p_n$ such that $r[B_n']\subseteq B_n$. It follows that $f(x)=\frac1n$ for all $x\in I$ such that $\left|x-\frac1n\right|<\epsilon_n$.

Now consider $x\in\left(\frac1{n+1},\frac1n\right)$. If $x<\frac1{n+1}+\epsilon_{n+1}$, then $f(x)<x$, and if $x>\frac1n-\epsilon_n$, then $f(x)>x$. By continuity of $f(x)-x$ there must be a point $x_n\in\left[\frac1{n+1}+\epsilon_{n+1},\frac1n-\epsilon_n\right]$ such that $f(x_n)=x_n$. This is possible, however, iff $r(\langle x_n,1\rangle)=\langle x_n,0\rangle$. For $n\in\Bbb Z^+$ let $q_n=\langle x_n,1\rangle$. Then

$$\begin{align*} &\langle q_n:n\in\Bbb Z^+\rangle\to\langle 0,1\rangle\;,\\ &\langle p_n:n\in\Bbb Z^+\rangle\to\langle 0,1\rangle\;,\\ &\langle r(q_n):n\in\Bbb Z^+\rangle\to\langle 0,0\rangle\;,\text{ and}\\ &\langle r(p_n):n\in\Bbb Z^+\rangle\to\langle 0,1\rangle\;, \end{align*}$$

which is impossible.

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