Continuous image of a connected set is connected

Statement

Let $X$ and $Y$ be topological spaces, let $f \colon X \to Y$ be continuous and let $S \subseteq X$ be connected. Then $f(S)$ is connected.

Remarks

Proof: if open sets $U$ and $V$ split $f(S)$, then the preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open and split $S$. Thus connectedness is a topological property. For $Y = \mathbb{R}$, this gives the intermediate value theorem. For example, the circle is connected, because it is the image of $[0, 1]$ under $t \mapsto (\cos 2\pi t, \sin 2\pi t)$.

Explanations

L1 human

A continuous map cannot break a space that is in one piece into separate pieces.