Let $f: X \to Y$ be a continuous function between topological spaces.
Denote for $x \in X, K(x)$ as the largest connected subset that contains $x$. Is it true that $f(K(x))$ is the largest connected set in $Y$ that contains $f(x)$?
Using continuity it is straightforward to see that $f(K(x)) \subseteq K(f(x))$, and if we assume that $f$ is a homeomorphism the other inclusion can be done in the same way. Can we prove the other inclusion only using that $f$ is continuous (and not bijective and/or continuous)?
Let $f : [0, 1) \cup [2, 3] \to [0, 2]$
$$f(x) = \begin{cases} x & \text{for } x \in [0, 1) \\ x-1 & \text{for } x \in [2, 3] \end{cases}$$
Then $f$ is a continuous bijection, but $f \big[ [2, 3] \big] = [1, 2] \subseteq [0, 2]$ is not maximally connected.