why $p^{-1}(W) \subset U ?$ why not $p^{-1}(W) = U ?$

Question

Let $p:\ X\to Y$ be a closed continuous surjective map such that $p^{-1}(\{y\})$ is compact for each $y\in Y$. Show that if $Y$ is compact, then $X$ is compact.

It is written

Let $U$ be an open set containing $p^{-1}(\{y\})$. Since $X-U$ is closed, $p(X-U)$ is closed in $Y$. Then $W=Y-p(X-U)$ is an open set that contains $y$. Since $(X-U)\cap p^{-1}(W)=\varnothing$, $p^{-1}(\{y\})\subset p^{-1}(W) \subset U$.

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My confusion :why $p^{-1}(W) \subset U ?$ why not $p^{-1}(W) = U ?$

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My thinking : $W=Y-p(X-U) \implies p^{-1}(W)=p^{-1}(Y) - p^{-1} p(X-U)$

since $p^{-1}(Y)=X \implies p^{-1}(W)=X - (X-U)=U$

Therefore $p^{-1}(W)=U$

Answer 1

$p^{-1}p(B) = B$ holds if and only if $p$ is injective.

Similarly, $pp^{-1}(B) = B$ if and only if $p$ is surjective.

Indeed, $p(U) = W$ means that $p(U)\subseteq W$ and $W\subseteq p(U)$. However, this does in general not imply that $U \subseteq p^{-1}(W)$.

Consider any non-injective map, say $f\colon \mathbb R\to \mathbb R,\ x\mapsto x^2$ as Randall suggested.

Clearly, $f(\mathbb R^+) = \mathbb R^+$. However, $f^{-1}(\mathbb R^+) =\mathbb R \not= \mathbb R^+$.