Let $(X,\tau_x)$ first counable space , $(Y,\tau_y)$, prove that pra image f of an open set in Y, contains open set

Question

Let $(X_1, \tau_1)$ be a first countable space, $(X_2,\tau_2)$ topological space, function $f\ :\ (X_1, \tau_1) \rightarrow (X_2,\tau_2)$ , V be any open set in $X_2$ prove that $f^{-1}(V)$ contains open set that is not an empty set. i'm stack in proving this.

Answer 1

It is not true.

In the first place $V$ can be empty, and in that case also $f^{-1}(V)$ is empty so that it cannot contain a non-empty set.

If - in order to avoid that situation - it is demanded that $V$ is not empty then still it is not true.

Let the first space be $\mathbb R$ equipped with usual topology (i.e. order topology).

Let the second space be $\mathbb R$ equipped with discrete topology.

Let $f$ be the identity function on $\mathbb R$.

Then the first space is first countable, $V=\{0\}$ is open in the second space and is not empty, but $f^{-1}(V)=\{0\}$ does not contain a non-empty set that is open in the first space.