Proposition

Let $f: X \rightarrow Y$ be a continuous map between topological spaces, and let $\mathcal{B}$ be a basis for the topology of $X$. Let $f(\mathcal{B})$ denote the collection $\{ f(B) : B \in \mathcal{B} \}$ of subsets of $Y$. Show that $f(\mathcal{B})$ is a basis for the topology of $Y$ if and only if $f$ is surjective and open.

Solution

Suppose that $f(\mathcal{B})$ is a basis for $Y$. Let $U \subset X$. Then $U = \bigcup B_{\alpha}$ for some $\{ B_{\alpha} \} \subset \mathcal{B}$. $f(U) = f(\bigcup B_{\alpha}) = \bigcup f(B_{\alpha})$. For each $\alpha$, $f(B_{\alpha})$ is a basis element for $Y$, so $\bigcup f(B_{\alpha})$ is open in $Y$. Therefore, $f$ is an open map.

$Y$ is open in $Y$, so there exists a subset $\{ f(B_{\alpha}) \} \subset f(\mathcal{B})$ such that $\bigcup f(B_{\alpha}) = Y$. Thus $Y = \bigcup f(B_{\alpha}) = f(\bigcup B_{\alpha}) \subset f(X) \subset Y$, so $f(X) = Y$.

Therefore, $f$ is both surjective and open.

On the contrary, suppose that $f$ is both surjective and open. We will show that $f(\mathcal{B})$ is a basis for $Y$. Let $V \subset Y$ be an open set and $y \in V$. Since $f$ is surjective, there exists an $x \in X$ such that $f(x) = y$. Since $f$ is continuous, $f^{-1}(V)$ is open in $X$. Since $f(x) = y \in V$, $x \in f^{-1}(V)$.

$\mathcal{B}$ is a basis for $X$, so there must exist a basis element $B \in \mathcal{B}$ such that $x \in B \subset f^{-1}(V)$. Then $f(x) \in f(B) \subset f(f^{-1}(V)) = V$. Therefore, $f(\mathcal{B})$ is a basis for $Y$ by Lemma 13.2 on P.80 (Munkres).