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 the composite mapping $g \circ f: T_1 \to T_3$ is continuous.

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

As $g$ is continuous, $g^{-1} \sqbrk U \in T_2$ is open in $T_2$.

As $f$ is continuous, $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in T_1$ is open in $T_1$.

By Inverse of Composite Bijection, $f^{-1} \sqbrk {g^{-1} \sqbrk U} = \paren {g \circ f}^{-1} \sqbrk U$.

Hence the result.