Composite of Continuous Mappings is Continuous

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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.


Corollary

Let $T_1, T_2, T_3$ each be one of:

metric spaces
the complex plane
the real number line


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.


Continuity at a Point

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

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

Let the mapping $g : T_2 \to T_3$ be continuous at $x$ at $\map f x$.

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


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.

$\blacksquare$


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