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Hatcher claims that there exists no homeomorphism $I \times I$ to $I \times \{0\} \cup A \times I$ where $A = \{0, 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \}$ on the basis of forward continuity, but he doesn't explain his logic fully. Instead, he just says that it can be attributed to "the bad structure of $(X,A)$ near $0$."

I've included a link to a picture that shows a proposed homeomorphism. It is a proposed deformation retraction $r: X \to A$ defined by radial projection from the point $(\frac{1}{n} + \frac{1}{2}(\frac{1}{n} - \frac{1}{n-1}), 1.5)$ to its corresponding section of the topological comb defined by $[\frac{1}{n}, \frac{1}{n-1}] \times \{0\} \cup \delta([\frac{1}{n}, \frac{1}{n-1}] \times I)$ for any $n \in \mathbb{N}$. I'm sure I'm not the first to think of this, but how exactly does this homeomorphism fail in forward continuity, bijectivity, or backward continuity?

Picture of proposed homeomorphism

And, in general, why is it that there can exist no homeomorphism from $I \times I$ to the topological comb?

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    $\begingroup$ Do you know what a homeomorphism is? The map you sketch is obviously very far from being injective, and it's easy to see that the two spaces are not homeomorphic. If you mean to ask about existence of a deformation retraction, that has been asked and answered on this site already. $\endgroup$ Commented Nov 11 at 22:02
  • $\begingroup$ A deformation retract is not a homeomorphism. $\endgroup$ Commented Nov 11 at 22:08
  • $\begingroup$ I get it: you want to answer quickly and stuff. But answers in the comments are discouraged, people; it's just not what they're for. Please stop. $\endgroup$ Commented Nov 12 at 0:16

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Let $K= I\times \{0\}\cup A\times I$. If $H\colon I^2\mapsto K$ is a homeomorphism, let $x_0=H^{-1}((2/3,0))$. Then $K\setminus \{(2/3,0)\}$ is not connected but $I^2\setminus \{x_0\}$ is connected. Contradiction.

Another way, using big guns: If $H\colon I^2\mapsto K$ is a homeomorphism then $H\colon (0,1)^2\rightarrow \mathbb{R}^2 $ is continuous and injective. By the Invariance of the domain we have that $H((0,1)^2)$ is a open subset of $\mathbb{R}^2$. That is impossible since $K$ has empty interior in $\mathbb{R}^2$. Of course this generalizes to: $I^2$ cannot be homeomorphic to a subset of $\mathbb{R}^2$ with empty interior in $\mathbb{R}^2$.

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