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Let $(X,\tau),(Y,\eta)$ be topological spaces. If $f:X\to Y$ is continuous, $f(X)=Y$ and $X$ is compact, then $Y$ is compact.

We want to show that for every open cover for $Y$ there exist a finite open subcover. So let $\{U_\alpha:\alpha\in I\}$ be ab open cover for $Y$. Thus $Y=\bigcup_{\alpha\in I} U_\alpha$

Now how can I use the hypothesis that $f$ is continuous?

I was thinking about this property of continuity: $\forall U\in\eta,f^{-1}(U)\in\tau$ but I don't know how to make a relation with the compactness of $X$.

As $X$ is compact, then $\forall U\subset\tau,X\subset\bigcup U$ and $\exists u_1,...,u_n\in U:U\subset\bigcup_iU_i$.

Could anyone help me please?

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    To use compactness of $X$, you need a collection of open sets which cover $X$. Hmm... I wonder where you're going to get one of those? All you have is a collection of open sets which cover $Y$. How can you turn open sets in $Y$ into open sets in $X$? Hmm... – Alex Kruckman Dec 09 '17 at 05:57

2 Answers2

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Consider the collection of sets $f^{-1}(U_{\alpha})$ for all $\alpha$.

They are open as $f$ is continuous and cover $X$ since $f$ is surjective.

Since $X$ is compact take a finite subcover $f^{-1}(U_{\alpha_1}), \ldots, f^{-1}(U_{\alpha_n})$.

What can you now deduce about $U_{\alpha_1}, \ldots, U_{\alpha_n}$?

Luke Peachey
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  • As $X$ is compact, $X=\bigcup_if^{-1}(U_{\alpha_i} )$. Therefore $Y=f(X)\subset f (\bigcup_{i} f^{-1}(U_{\alpha_i}))$

    $=\bigcup_i U_{\alpha_i}$ ?

     –  Dec 09 '17 at 06:24
  • Yes, almost correct, the only issue is that you need to write the index of the union clearly to indicate that is the finite union, if not, it is confusing if you are doing uncountably union or what. – user284331 Dec 09 '17 at 06:45
  • @user284331 ok, thanks. –  Dec 09 '17 at 07:31
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Write $X=\displaystyle\bigcup_{\alpha\in I}f^{-1}(U_{\alpha})$. Continuity is for $f^{-1}(U_{\alpha})$.

user284331
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  • Related:https://math.stackexchange.com/questions/323610/the-definition-of-continuous-function – Peter Szilas Dec 09 '17 at 07:41
  • Can you explain how the hypothesis of surjective is applied? I can see that $f^{-1}(U_{\alpha}),\forall \alpha\in I$ are all in $X$. When it says $f^{-1}(U_{\alpha}),\forall \alpha\in I$ it means that $f^{-1}$ is evaluated in every set of Y? –  Dec 09 '17 at 20:12
  • in other words, after the evaluation in $f^{-1}$, Y will be 'empty' since all it's elements have been evaluated? –  Dec 09 '17 at 20:14
  • In your below comment, you have written $f\left(\displaystyle\bigcup_{i}f^{-1}(U_{\alpha_{i}})\right)=\displaystyle\bigcup_{i}f(f^{-1}(U_{\alpha_{i}}))=\displaystyle\bigcup_{i}U_{\alpha_{i}}$. Surjective: $f(f^{-1}(A))=A$. – user284331 Dec 09 '17 at 20:16
  • Sorry I don't quite follow your second question. – user284331 Dec 09 '17 at 20:17
  • but in the answer below, he says that as $f$ is surjective then $f^{-1}(U_\alpha)$ is a cover for $X$, why? –  Dec 09 '17 at 20:26
  • I don't quite agree. For example, if $f$ is not surjective, if you put $f^{-1}(Y)$, since the codomain of $f$ is $Y$, one must have $X=f^{-1}(Y)$. Surjectivity is used when you have found a finite cover of $X$, and the crucial equality that $Y=f(X)$. – user284331 Dec 09 '17 at 20:29
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    Sorry, just now I should not say that surjectivity is used for $f(f^{-1}(A))=A$ (although this is correct by the property of surjectivity, but this does not help that much in the proof, because we always have $f(f^{-1}(A))\subseteq A$). Surjectivity is for $Y=f(X)$, in some sense, this is the crucial step in the proof. – user284331 Dec 09 '17 at 20:30
  • Then how do you assure that $X=\displaystyle\bigcup_{\alpha\in I}f^{-1}(U_{\alpha})$? –  Dec 09 '17 at 20:35
  • We have $X=f^{-1}(Y)=f^{-1}\left(\displaystyle\bigcup_{\alpha}U_{\alpha}\right)=\displaystyle\bigcup_{\alpha}f^{-1}(U_{\alpha})$. The tip is, $f^{-1}\left(\displaystyle\bigcup_{\gamma}A_{\gamma}\right)=\displaystyle\bigcup_{\gamma}f^{-1}(A_{\gamma})$ is always a true formula. – user284331 Dec 09 '17 at 20:36
  • oh you're right, I forgot to consider the open cover for $Y=\cup U_\alpha$, now it's clear, thanks again. –  Dec 09 '17 at 20:43