Does continuity of a map $f$ imply: $f^{-1}$ maps subbasis onto subbasis?

I am stuck with an exercise, maybe you can help me out here.

Let $(X,\tau)$ be a topological space with subbasis $\mathcal{S}$ and let $f: X \longrightarrow X$ be a surjective map.

1. Show that $f^{-1}(S) \in \mathcal{S}$ for all $S \in \mathcal{S}$ implies that $f$ is continuous.
2. Does the converse hold?

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1. $\quad$ Any open set can be written as a union of basis elements. A basis element is a finite intersection of elements in the subbasis. By assumption the preimage of a finite intersection of subbasis-elements is a finite intersection of subbasis-elements again and therefore a basis element. Consequently, the preimage of an open set is the union of basis elements - which again is an open set. $\quad \square$

Question: Do I even need surjectivity here?

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2. $\quad$ I am stuck here. Neither am I able to give a counter-example nor was I able to prove that the converse indeed holds. I somehow have the feeling that the converse does not hold.

Question: Do you have any tips on how to tackle this or can you provide an instructional counter-example?

Solution 1:

Question 1: no surjectivity is needed: the fact that subbasic sets have open preimages is a simple necessary and sufficient condition for continuity in general, because $f^{-1}$ on sets preserves set operations.

The converse need not hold: take $X=\Bbb R$ with the standard subbase

$$\mathcal{S} = \{(-\infty,a),(a,+\infty)\mid a \in \Bbb R\}$$ and $f(x)=x\sin x$ as an example.