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Let $U$ be a simply connected open set of $\mathbb{R}^2$, and $\phi:[0,1] \rightarrow U$ a cross-cut, that is an injective continuous map such that $\phi((0,1)) \subset U$, while $\phi(0) \in \mathbb{R}^2 \backslash U$ and $\phi(1) \in \mathbb{R}^2 \backslash U$.

Then $U \backslash \phi((0,1))$ has exactly two connected components which are both simply connected (see e.g. Newman, Elements of the Topology of Plane Sets of Points, Chapter VI, Theorem (5.1)), say $D_1$ and $D_2$.

Is $D_{i} \cup \phi((0,1))$ homeomorphic to the half plane $H= \{(x,y) \in \mathbb{R}^2 : x \geq 0 \}$?

Thank you very much for your attention.

NOTE. Since each $D_i$ is a simply connected open subset of $\mathbb{R}^2$, we know that it is homeomorphic to $\mathbb{R}^2$ (see e.g. Newman, Elements of the Topology of Plane Sets of Points, Chapter VI, Theorem (6.4)).

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  • $\begingroup$ Not to nitpick, but isn't it redundant to say "a connected set which is simply connected," since the latter implies the former? $\endgroup$
    – Math1000
    Commented Jan 29, 2017 at 20:24
  • $\begingroup$ My thought is that they would not be homeomorphic due to the boundary of $D_i\cup\phi((0,1))$ being disconnected, while the boundary of $H$ is. $\endgroup$
    – Math1000
    Commented Jan 29, 2017 at 20:33
  • $\begingroup$ @Math1000: thank you for your observation, I smoothed the wording. $\endgroup$ Commented Jan 29, 2017 at 22:00

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