Perfect map: If $Y$ is compact, then $X$ is compact?

Question

I am trying to understand this proof and everything was pretty well until the last part of the second paragraph.

...Since $Y$ is compact, we can choose a finite subcover $W_{y_1},...,W_{y_m}$ covering $Y$. Then the collection $V_{y_1},...,V_{y_m}$ covers $X$...

I am having difficulties understanding the bold text.

I have done this to try to understand it. $$f(V_y)=\bigcup_{j=1}^mf(U_{i_j})\supset\bigcup_{i=1}^m W_{y_i}=Y=f(X)...(1)$$ $$X=\bigcup_{i=1}^mV_{y_i}...(2)$$

My question is how do you pass form (1) to (2) ?

Answer 1

I think that in the second part the phrase:

"the sets $p(V_{y})$ each contain some open neighborhood $W_y$ of $y$..."

must be replaced by the stronger statement:

"for every set $V_{y}$ we can find an open neighborhood of $W_y$ of $y$ such that $p^{-1}(W_{y})\subseteq V_{y}$...".

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Let $x\in X$.

Then $p(x)\in Y=\bigcup_{i=1}^m W_{y_i}$ so $p(x)\in W_{y_i}$ for some $i\in\{1,\dots,m\}$.

Then $x\in p^{-1}(W_{y_i})\subseteq V_{y_i}$ because the $W_{y_i}$ are chosen in such a way that $p^{-1}(W_{y_i})\subseteq V_{y_i}$.

The fact that is is possible has been proved in the first part.

Proved is now that the sets $V_{y_1},\dots V_{y_m}$ cover $X$.

Answer 2

$X=p^{-1}(Y)=\cup_{i=1}^{i=m}p^{-1}(W_{y_i})$ since $p^{-1}(W_{y_i})\subset V_{y_i}$, you deduce that $X= \cup_{i=1}^{i=m}V_{y_i}$.