Composite of Quotient Mappings in Topology is Quotient Mapping

Theorem
Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$, $T_3 = \struct {S_3, \tau_3}$ be topological spaces.

Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be quotient mappings.

Then $g \circ f : S_1 \to S_3$ is a quotient mapping.

Proof
Composite of Surjections is Surjection shows that $g \circ f$ is surjective.

Composite of Continuous Mappings is Continuous shows that $g \circ f$ is continuous.

Let $U \subseteq S_3$ such that $\paren {g \circ f}^{-1} \sqbrk U$ is open in $T_1$.

By definition of quotient mapping, $f \sqbrk { \paren {g \circ f}^{-1} \sqbrk U }$ is open in $T_2$.

By definition of quotient mapping, $g \circ f \sqbrk { \paren {g \circ f}^{-1} \sqbrk U }$ is open in $T_3$.

Image of Preimage under Mapping/Corollary shows that $g \circ f \sqbrk { \paren {g \circ f}^{-1} \sqbrk U } = U$.

It follows that $U$ is open in $T_3$.

By definition of quotient mapping, it follows that $g \circ f$ is a quotient mapping.