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Problem of the Week #228 - Oct 11, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Call an $S$-space over a topological space $B$ a pair $(E,p)$ where $E$ is a topological space and $p$ is a local homeomorphism from $E$ into $B$. A morphism of $S$-spaces $(E_1,p_1)$, $(E_2,p_2)$ over $B$ is a continuous mapping $\phi : E_1 \to E_2$ such that $p_1 = p_2 \circ \phi$. Show that if $\phi$ is a morphism of $S$-spaces, then $\phi$ is a local homeomorphism if and only if $\phi$ is an open mapping.


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
No one solved this week's problem. You can read my solution below.


Let $\phi : (E_1,p_1) \to (E_2,p_2)$ be a morphism of $S$-spaces over $B$.

Suppose $\phi$ is a local homeomorphism. Fix an open set $U$ in $E_1$ and take $y\in \phi(U)$. There is an $x\in U$ such that $\phi(x) = y$; since $\phi$ is a local homeomorphism, there exists an open neighborhood $G$ of $x$ such that $\phi|G : G \to \phi(G)$ is a homeomorphism. Then $\phi(G\cap U)$ is an open neighborhood of $y$ contained in $\phi(U)$. Thus $\phi(U)$ is open, and consequently $\phi$ is an open mapping.

Conversely, assume $\phi$ is an open mapping. Let $x\in E_1$. Since $p_1$ is a local homeomorphism, there exists an open neighborhood $U$ of $x$ such that $p_1 | U : U \to p_1(U)$ is a homeomorphism. Since $p_2$ is a local homeomorphism, there is an open neighborhood $V$ of $\phi(x)$ such that $p_2 | V \to p_2(V)$ is a homeomorphism. The set $G = U \cap p_2^{-1}(V)$ is an open neighborhood of $x$ for which the composition $G \xrightarrow{\phi|G} \phi(G) \xrightarrow{p_2|G} p_1(G)$ is $p_1|G$. Since $p_1|G$ and $p_2|G$ are homeomorphisms, so is $\phi|G$. Hence, $\phi$ is a local homeomorphism.
 
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