Image of an intersection is the intersection of the image

Question

I'm studying the proof that path-connectedness implies connectedness. However there is one assertion that I don't understand.

Let $f:[0,1] \rightarrow X$ be a continuous mapping where $X$ is a metric space. Let $U$ and $V$ be two disjoint subsets of X such that $X = U \cup V$.

I don't understand why $f^{-1}(U) \cap f^{-1}(V) = \emptyset $.

Is it because:

$$ \emptyset = f^{-1}(\emptyset) = f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$$

That makes sense but I thought the last equality $f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$ only holds iif $f^{-1}$ is injective. Does that mean $f^{-1}$ is injective or am I missing something?

Or is it because $f$ is continuous?

For a reference to the proof: https://topospaces.subwiki.org/wiki/Path-connected_implies_connected

Answer 1

$f^{-1}[U \cap V] = f^{-1}[U] \cap f^{-1}[V]$ always holds. The function $f$ need not be 1-1 or onto etc. You’re confused with $f[U \cap V] = f[U] \cap f[V]$ which does only hold if $f$ is injective. Forward and inverse images behave quite differently.

Answer 2

It $t\in f^{-1}(U)\cap f^{-1}(V)$, then $f(t)\in U\cap V$. But we are assuming that $U$ and $V$ are disjoint.