Let $f : X → Y$ be a map between two topological spaces $X$ and $Y$ , and $S_Y ⊆ 2^Y$ a sub-basis for the topology of $Y$ . Show that $f$ is continuous if and only if for every $B ∈ S_Y$ , the set $f^{−1}(B)$ is open in $X$.
My attempt:
Definition: a function $f: X→Y$ is continuous if for every open $U \in Y$, we have $f^{-1}(U)$ is open in $X$.
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$=>$: If $B \in S_Y$, then $B$ is open in $Y$. Because $f$ is continuous, $f^{-1}(B)$ is open, by definition of continuity.
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$<=$: Let $V⊆Y$ be an open subset, and since $S_Y$ is a sub-basis, then $Y = U_{B \in S}B$, so $V ⊆ U_{B \in S}B$, so we can write $V$ as $V = U(\cap_{B \in S}B)$. Hence, $f^{-1}(V)= f^{-1}(U(\cap_{B \in S}B)) = Uf^{-1}(\cap_{B \in S}B)$, and since $B$ is open, then so as $\cap_{B \in S} B$, thus $Uf^{-1}(\cap_{B \in S}B)$ is open, which implies that $f^{-1}(V)$ is open, hence $f$ is continuous.
Is my attempt correct? Any help please?
Best Answer
A straightforward proof of $(\Leftarrow)$ is the following. Let $V$ be any open subset of $Y$ and $y\in Y$ be ainy point. There exists a finite subfamily $\mathcal B_y$ of $S_Y$ such that $y\in\bigcap \mathcal B_y\subset V$. Then $f^{-1}(y)\subset f^{-1}\left(\bigcap \mathcal B_y\right)\subset f^{-1}(V)$. Thus a set
$$f^{-1}(V)=\bigcup_{y\in Y} f^{-1}\left(\bigcap \mathcal B_y\right)= \bigcup_{y\in Y} \bigcap_{U\in \mathcal B_y} f^{-1}(U)$$ is open.