Continuous Image of Path-Connected Set is Path-Connected
Theorem
Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be path-connected in $\struct {T_1, \tau_1}$.
Then $f \sqbrk S$ is path-connected in $\struct {T_2, \tau_2}$.
Metric Space
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 $f \sqbrk S$ is a path-connected subspace of $M_2$.
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 $\struct {T_1, \tau_1}$.
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.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $6.3$: Path-connectedness (in passing)