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I have two seemingly simple questions in topology but I have no idea how to answer them.

1) Let $\varphi:X\to Y$ be a continuous map such that $\varphi^0:X\to\mathrm{Im}(\varphi)$ is a homemorphism. Assume we have a representation $\varphi=\psi\circ\chi$, where $\chi:X\to Z$ is a bijective continuous map, and $\psi:Z\to Y$ is a continuous map. How to prove that $\chi:X\to Z$ is a homeomorphism? Maps for which such an implication holds are called extreme monomorphisms.

2) Let $\varphi:X\to Y$ be a continuous map and topology of $Y$ is the strongest topology for which $\varphi$ is continuous. Assume we have a representation $\varphi=\chi\circ\psi$, where $\chi:Z\to Y$ is a bijective continuous map, and $\psi:X\to Z$ is a continuous map. How to prove that $\chi:X\to Z$ is a homeomorphism? Maps for which such an implication holds are called extreme epimorphisms.

Thanks for your help.

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Are you certain that these are even true? –  tomasz Aug 17 '12 at 21:20
    
I found this statements in one particular textbook. –  userNaN Aug 17 '12 at 21:22
    
@Norbert Could you please indicate which textbook? –  magma Jan 17 at 9:43
    

1 Answer 1

up vote 2 down vote accepted

(1) It suffices to prove that $\chi$ is an open map. Suppose that $V$ is an open set in $X$ such that $A=\chi[V]$ is not open in $Z$. Let $B=\psi[A]\subseteq\varphi[X]$. Clearly $V=(\psi\circ\chi)^{-1}[B]=\varphi^{-1}[B]=(\varphi^0)^{-1}[B]$, and $\varphi^{0}$ is a homeomorphism, so $B=\varphi^0[V]$ must be open in $\varphi[X]$. It follows that $\psi^{-1}[B]$ is open in $Z$. If $\psi^{-1}[B]=A$, we’re done. If not, there must be some $z\in\psi^{-1}[B]\setminus A$. Let $x$ be the unique element of $X$ such that $\chi(x)=z$. Then $x\notin V$, but $\varphi^0(x)=(\psi\circ\chi)(x)=\psi(z)\in B=\varphi^0[V]$, contradicting the assumption that $\varphi^0$ is bijective.

(2) Assuming that you actually want to prove that $\chi:Z\to Y$ is a homeomorphism, you can use the same idea. Again it suffices to show that $\chi$ is open, so let $[V]$ be an open subset of $Z$, let $A=\chi[V]$, and let $W=\psi^{-1}[V]$, an open set in $X$. Clearly $A=\chi[V]=(\chi\circ\psi)[W]=\varphi[W]$. Let $\tau$ be the topology on $Y$. If $A\notin\tau$, the topology on $Y$ generated by $\tau\cup\{A\}$ would be a strictly stronger topology with respect to which $\varphi$ was continuous.

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Thanks for your answer. Could you explain why $\psi^{-1}[B]$ is open? –  userNaN Aug 17 '12 at 22:10
    
@Norbert: Because $\psi$ is continuous. –  Brian M. Scott Aug 17 '12 at 22:25
    
But $B$ is open in $\varphi[X]$ not in $Y$ –  userNaN Aug 17 '12 at 22:27
    
@Norbert: It makes no difference: $\psi[Z]=\varphi[X]$. –  Brian M. Scott Aug 17 '12 at 22:31
    
Thnanks, now it is clear to me. –  userNaN Aug 17 '12 at 22:33

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