# Continuous Image of Path-Connected Set is Path-Connected

## Theorem

Let $M_1, M_2$ be metric spaces whose metrics are $d_1, d_2$ respectively.

Let $f: M_1 \to M_2$ be a continuous mapping.

Let $S \subseteq M_1$ be a path-connected subspace of $M_1$.

Then $f \left({S}\right)$ is a a path-connected subspace of $M_2$.

## Proof

Let $f \left({s}\right), f \left({s'}\right) \in f \left({S}\right)$, for some $s, s' \in S$.

Let $\mathbb I$ denote the closed unit interval:

$\mathbb I = \left[{0 \,.\,.\, 1}\right]$

Let $p: \mathbb I \to S$ be a continuous mapping such that:

$p \left({0}\right) = s, p \left({1}\right) = s'$

Such a $p$ exists since $S$ is a path-connected subspace of $M_1$.

Now define $q: \mathbb I \to f \left({S}\right)$ by:

$q \left({t}\right) := f \circ p \left({t}\right)$

Then $q$ is a continuous mapping, being the composition of such, and:

$q \left({0}\right) = f \left({p \left({0}\right)}\right) = f \left({s}\right)$
$q \left({1}\right) = f \left({p \left({1}\right)}\right) = f \left({s'}\right)$

Thus $f \left({s}\right)$ and $f \left({s'}\right)$ are connected by a path in $f \left({S}\right)$.

Since these points were arbitrary in $f \left({S}\right)$, the result follows.

$\blacksquare$