Continuity of Composite Mapping

Theorem

Let $T_1, T_2, T_3$ be topological spaces.

Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be continuous mappings.

Then $g \circ f: T_1 \to T_3$ is continuous.

Proof

Let $U \in T_3$ be open in $T_3$.

As $g$ is continuous, $g^{-1}\left({U}\right) \in T_2$ is open in $T_2$.

As $f$ is continuous, $f^{-1} \left({g^{-1}\left({U}\right)}\right) \in T_1$ is open in $T_1$.

By Inverse of Composite Bijection, $f^{-1} \left({g^{-1}\left({U}\right)}\right) = \left({g \circ f}\right)^{-1} \left({U}\right)$.

Hence the result.

$\blacksquare$

Corollary

Because:

• Real Number Line is Metric Space
• Complex Plane is Metric Space
• Metric Induces a Topology

it follows that there is no need to prove continuity of composites of real functions or complex functions separately.

Sources

• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{III}$
• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 1$: Functions