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 $X$.
$=>$: if $f$ is continuous then $f$ is continuous at each $x \in X$, so for every $x \in X$, $f(x) = y \in Y$. And since $S_Y$ is a sub-basis for the topology of $Y$, then $Y = U_{B \in S} B$, so $y = f(x) \in U_{B \in S} B$. Hence $x \in f^{-1}(U_{B \in S} B)$ which is open in $X$ for every open $B ⊆ Y$.
$<=$: 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?
You have greatly over-thought the forward direction. In fact, I would say that that is the trivial direction. Here is a much shorter proof of it:
If $B \in S_Y$, then $B$ is open in $Y$. Because $f$ is continuous, $f^{-1}(B)$ is open, by definition of continuity.
Now, as for the reverse, you have the right idea, but have become a little muddled in it. What you want to show is that if $U$ is any arbitrary open subset of $Y$, then $f^{-1}(U)$ is open. To do this, you can use the definition of a sub-basis to "build" $U$ out of elements of $S_Y$, and then prove that $f^{-1}$ is "nicely behaved" with this construction. Of course, those two phrases I put in quotes cannot simply be glossed over, and need to be worked out in full.