# If $f:X\to Y$ is continuous, $X$ is compact and $f(X)=Y,$ then $Y$ is compact.

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 $$\exists u_1,...,u_n\in U:U\subset\bigcup_iU_i$$.

• 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 9 '17 at 5:57

Write $X=\displaystyle\bigcup_{\alpha\in I}f^{-1}(U_{\alpha})$. Continuity is for $f^{-1}(U_{\alpha})$.

• – Peter Szilas Dec 9 '17 at 7:41
• 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? – user486983 Dec 9 '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? – user486983 Dec 9 '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 9 '17 at 20:16
• Sorry I don't quite follow your second question. – user284331 Dec 9 '17 at 20:17

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 $U_{\alpha_1}, \ldots, U_{\alpha_n}$?

• As $X$ is compact, $X=\bigcup_if^{-1}(U_{\alpha_i} )$. Therefore $Y=f(X)\subset f (\bigcup_{i} f^{-1}(U_{\alpha_i}))$ $=\bigcup_i U_{\alpha_i}$ ? – user486983 Dec 9 '17 at 6:24
• 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 9 '17 at 6:45
• @user284331 ok, thanks. – user486983 Dec 9 '17 at 7:31