Composite of Continuous Mappings is Continuous

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, $$f^{-1} \left({g^{-1}\left({U}\right)}\right) = g \circ f \left({U}\right)$$.

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.