Proving open map with a subbasis

Let $(𝑋, T 𝑋)$ and $(𝑌, T 𝑌)$ each be a topological space, let $S$ be a subbasis of $T 𝑋$ , and let $𝑓 : 𝑋 \to 𝑌$ . Then $𝑓$ is an open map iff the image $𝑓 (𝑉)$ is open for every subbasic open set $𝑉 \in S$ . #m/thm/topology

Proof

Clearly if $𝑓$ is open the image of every $𝑉 \in S$ is open. For the converse, first consider the completed basis $B$ . Let $𝑉 \in B$ , implying there exists a finite sequence $(𝑆 𝑖) 𝑛 𝑖 = 1 \in S$ such that $𝑉 = ⋂ 𝑛 𝑖 = 1 𝑆 𝑖$ . Then

𝑓 (𝑉) = 𝑓 (𝑛 ⋂ 𝑖 = 1 𝑆 𝑖) = 𝑛 ⋂ 𝑖 = 1 𝑓 (𝑆 𝑖)

which is the finite intersection of open sets and is thus open. Hence $𝑓 (𝑉)$ is open for all $𝑉 \in B$ . Now consider the whole generated topology $T 𝑋$ . Let $𝑉 \in T 𝑋$ , implying there exist $(𝑆 𝑖) 𝑖 \in 𝐼 \in B$ such that $𝑉 = ⋃ 𝑖 \in 𝐼 𝑆 𝑖$ . Then

𝑓 (𝑉) = 𝑓 (⋃ 𝑖 \in 𝐼 𝑆 𝑖) = ⋃ 𝑖 \in 𝐼 𝑓 (𝑆 𝑖)

which is the union of open sets and thus open. Hence the image of every open set is open, wherefore $𝑓$ is open.

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