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Every retract of a moscow space is a moscow space.

I tried to do it but its not true,

Let $A$ is a retract subspace of a moscow space $X$ ($\rho$ is retraction of $X$ onto $A$) and $U$ be an open subset of $A$, then $\rho^{-1}(U)$ is an open subset of $X$. since $X$ is a moscow space then for $a\in \overline{\rho^{-1}(U)}$ there exists a $G_\delta$-set, $P$ such that $a\in P\subset \overline{\rho^{-1}(U)} \subset \rho^{-1}(\overline{U})$.

$\rho(a)\in \rho(P)\subset \overline{U}$ but $\rho(P)$ maybe isnt $G_\delta$-set in $A$.

how can we proof this proposition?

Thanks.

TXC
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1 Answers1

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$\newcommand{\cl}{\operatorname{cl}}$Let $V=\rho^{-1}[U]$. Since $\rho$ is a retraction, $A\cap V=U$, and therefore $\cl_A U\subseteq A\cap\cl_X V$. On the other hand, $\rho[\cl_X V]\subseteq\cl_A U$, since $\rho$ is continuous, so $A\cap\cl_X V\subseteq\rho[\cl_X V]\subseteq\cl_A U$. It follows that $\cl_A U=A\cap\cl_X V$. If $G$ is any $G_\delta$-subset of $\cl_X V$, $A\cap G$ is a $G_\delta$-subset of $A\cap\cl_X V=\cl_A U$ in $A$, so $\cl_A U$ is a union of $G_\delta$-sets, and $A$ is a Moscow space.

Brian M. Scott
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