Continuous Image of Connected Space is Connected/Proof 1

Theorem

Let $T_1$ and $T_2$ be topological spaces.

Let $S_1 \subseteq T_1$ be connected.

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

Then the image $f \sqbrk {S_1}$ is connected.

Proof

Let $i: f \sqbrk {T_1} \to T_2$ be the inclusion mapping.

Let $g: T_1 \to f \sqbrk {T_1}$ be the surjective restriction of $f$.

Then $f = i \circ g$.

Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous.

We use a Proof by Contradiction.

Suppose that $A \mid B$ is a partition of $f \sqbrk {T_1}$.

Then it follows that $g^{-1} \sqbrk A \mid g^{-1} \sqbrk B$ is a partition of $T_1$.

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$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $6.2$: Connectedness: Proposition $6.2.10$
• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products