# 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.

### Corollary

Let $T_1, T_2, T_3$ be either:

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.

### Continuity at a Point

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

Let the function $f : T_1 \to T_2$ be continuous at $x$.

Let the function $g : T_2 \to T_3$ be continuous at $f\left({x}\right)$.

Then the function $g \circ f : T_1 \to T_3$ is continuous at $x$.

## 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.

$\blacksquare$