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$\varphi\colon M\to N$ continuous surjective and closed. Then $f\colon N\to P$ continuous iff $f\circ\varphi\colon M\to P$ is continuous. (Topological spaces)

I think that this proposition is true like I noted in $\varphi\colon M\to N$ continuous and open. Then $f$ continuous iff $f\circ\varphi$ continuous. (My commentary in Added (2)) which is true if we put $\varphi$ open instead of closed (in this case the proof is straightforward).

I tried a proof to this harder fact but is so large. I was put this as an possible answer to this question.

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up vote 2 down vote accepted

Here's a way of doing it:

Now, let $f\colon N\to P$, $\varphi\colon M\to N$ and assume that $\varphi$ is both continuous, onto, and closed. We assume that $f\circ\varphi$ is continuous, and we want to show that $f$ is continuous. Is suffices to show that the inverse image (under $f$) of a closed set is closed.

Let $C\subseteq P$ be closed. Since $f\circ \varphi$ is continuous, then $(f\circ\varphi)^{-1}(C)$ is closed. Therefore, $\varphi((f\circ\varphi)^{-1}(C))$ is a closed subset of $N$.

I claim that $\varphi((f\circ\varphi)^{-1}(C) = f^{-1}(C)$. Indeed, note that $(f\circ\varphi)^{-1}(C) = \varphi^{-1}(f^{-1}(C))$, and that since $\varphi$ is onto, we have $\varphi(\varphi^{-1}(B)) = B$ for all $B\subseteq N$.

Thus, $f^{-1}(C) = \varphi((f\circ\varphi)^{-1})(C))$, which is closed, so $f$ is continuous (since inverse image of closed subsets are closed). $\Box$

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Thanks!! My proof was ridiculously long and complicated. This exercise don't looks hard and must have a simple answer. My first attempt was using closed sets and preimages but when I got $\varphi[(f\circ\varphi)^{-1}[C]]$ I didn't think that would be useful and never think that be equal to preimage by $f$ of $C$. – Gastón Burrull Jun 20 '12 at 2:12
Now is better!! The equality follow easier! – Gastón Burrull Jun 20 '12 at 2:38

I think that I have the answer as I said in my question but I'm using many facts and my proof turns so large and I think sometimes that a large proof maybe is a wrong proof but this is my attempt:

The direction $\implies$ is trivial.

For $\Longleftarrow$ we want to prove continuity of $f\colon N\to P$ in some fixed point $a\in N$. Let $V\subset P$ some open set containing $f(a)$. We want to find an open subset $W\subset N$ such that $a\in W$ and $f[W]\subset V$

We note that $\varphi^{-1}[\{a\}]$ is not empty by surjectivity of $\varphi$, if $b\in\varphi^{-1}[\{a\}]$ we have $(f\circ\varphi)(b)=f(a)$ and by continuity of $f\circ\varphi$ in $b$ there exist an open set $U_b\subset M$ such that $(f\circ\varphi)[U_b]\subset V$. Repeating the same with each $b\in\varphi^{-1}[\{a\}]$ we obtain a collection of open sets such union $A=\bigcup_{b\in\varphi^{-1}[\{a\}]} U_b$ is an open set in $M$ such that $(f\circ\varphi)[A]\subset V$ and $\varphi^{-1}[\{a\}]\subset A$. Then is time to use the following Lemma:

Lemma: $f\colon M\to N$ (topological spaces) is closed if and only if for all $y\in N$ and all open sets $V\supset f^{-1}\left(\{y\}\right)$ in $M$ there exists an open set $U$ in $N$ containing $y$ such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$.

Proof: This Lemma was proved in $f$ closed iff $y\in N$ and open $V\supset f^{-1}\left(\{y\}\right)$ exists $U$ open such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$ $\square$.

By lemma, since $A$ is an open set containing $\varphi^{-1}[\{a\}]$ and $\varphi$ is closed there exists an open set $a\in B\subset N$ such that $\varphi^{-1}[B]\subset A$ therefore $$(f\circ\varphi)[\varphi^{-1}[B]]\subset (f\circ \varphi)[A]\subset V,$$ but by surjectivity of $\varphi$ we have that $\varphi[\varphi^{-1}[B]]=B$ then $$f[B]\subset V.\square$$

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Looks right on first read, but it seems a bit too complicated; I think the expected argument is to pull back a closed subset of $P$ all the way to $M$, and then push it forward with $\varphi$ to conclude that inverse image under $f$ of closed sets is closed. – Arturo Magidin Jun 20 '12 at 0:28

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