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Let $X$ be the subset $A \cup B$ of $\mathbb{R}^2$, where $$A = \{(x,y) \in \mathbb{R}^2 \mid x \le 0\text{ or }y \le 0\}$$ and $$B = \{(x,y) \in \mathbb{R}^2 \mid x \lt 1\text{ and }y \lt 1\}.$$

Equip $\mathbb{R}^2$ with the standard product topology and equip $X$ with the induced subspace topology. Find a continuous bijection $f: X \to \mathbb{R}^2$ that is not a homeomorphism.

I don't really know what to do for this question, other than I think it needs to be a piecewise function. I know it can't simply be a translation of $B$ along the $x$ and $y$ axes, since that would have a continuous inverse function and hence be a homeomorphism. I think it somehow has to do with a reciprocal function and the fact that the $x = 0$ and $y = 0$ are a part of $A$ but are not contained in open intervals.

Lääne-Viru
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emseon
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    http://math.stackexchange.com/questions/378717/finding-counterexamples-bijective-continuous-functions-that-are-not-homeomorphi may be of some help.. –  Jan 01 '15 at 13:46
  • The obvious translation is the correct answer. Look at points along where $A$ and $B$ meet to see that the inverse is not continuous. – Arthur Jan 01 '15 at 14:12

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Hint: Prove the quotient of $A\cup\overline B\setminus\{(1,1)\}$ given by $(t,1)\sim(0,\frac1{1-t})$ and $(1,t)\sim(\frac1{1-t},0)$, where $t\in(0,1)$ is homeomorphic to $\mathbb R^2$.

user2345215
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