Continuous Image of Path-Connected Set is Path-Connected

Theorem
Let $\paren{ T_1, \tau_1 }, \paren{ T_2, \tau_2 }$ be topological spaces.

Let $f: T_1 \to T_2$ be a continuous mapping.

Let $S \subseteq T_1$ be a subset of $T_1$.

Suppose $S$ is path-connected in $\paren{ T_1, \tau_1 }$.

Then $f \sqbrk S$ is path-connected in $\paren{ T_2, \tau_2 }$.

Proof
Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.

Let $\mathbb I$ denote the closed unit interval:
 * $\mathbb I = \closedint 0 1$

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


 * $\map p 0 = s, \map p 1 = s'$

Such a $p$ exists since $S$ is path-connected in $\paren{ T_1, \tau_1}$.

Now define $q: \mathbb I \to f \sqbrk S$ by:


 * $\map q t := f \circ \map p t$

Composite of Continuous Mappings is Continuous shows that $q$ is a continuous mapping, and:


 * $\map q 0 = \map f {\map p 0} = \map f s$
 * $\map q 1 = \map f {\map p 1} = \map f {s'}$

Thus $\map f s$ and $\map f {s'}$ are connected by a path in $f \sqbrk S$.

Since these points were arbitrary in $f \sqbrk S$, the result follows.