If $f$ is atomic then int $(f (U)) \neq \emptyset$ (interior). True or False?

Question

Let $f:X \to Y$ be a continuous function between continua. If $f$ is atomic then int $(f (U)) \neq \emptyset$ (interior).

I don't know if this conjecture is true. Before presenting my attempt, I will recall some definitions.

• A map $f : X \to Y$ between continua is said to be atomic if it is surjective and for every subcontinuum $K$ of $X$ such that $f (K)$ is nondegenerate, we have that $K = f ^{−1}(f (K))$.
• A map $f : X \to Y$ between continua is atomic if and only if $f ^{−1}(y)$ is a terminal subcontinuum in $X$ for each $y \in Y$ .

My attempt:

I first demonstrated the following.

Lemma 1: Let $f:X \to Y$ be a continuous function between continua. If $U$ is an open of $X$ and $f^{-1} (y) \subseteq U$ for some $y \in Y$, then $y \in $ int$f(U)$.

Now if i will start my attempt

Let $U$ be an open of $ X $. Let $ L $ be a nondegenerate continuum of $ X $ such that $L \subseteq U$.

1. If $|f(L)|>1$.

Let $y \in f(L)$, then $f^{-1}(y) \subseteq f^{-1}(f(L))$. Since $f$ is atomic then $L =f^{-1}(f(L))$, thus $f^{-1}(y) \subseteq f^{-1}(f(L))=L \subseteq U$. Therefore, by Lemma 1, $y \in $ int$f(U)$.

2. If $|f(L)|=1$.

This case I have not been able to prove it.

Any ideas or a counterexample? Thank you so much.