Continuous Open Surjective Mapping is Quotient Mapping

Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: S_1 \to S_2$ be a continuous open surjective mapping.

Then $f$ is a quotient mapping.

Proof

Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.

By definition of open mapping:

$f \sqbrk {f^{-1} \sqbrk U}$ is open in $T_2$

From Image of Preimage under Mapping: Corollary:

$f \sqbrk {f^{-1} \sqbrk U} = U$

It follows that $U$ is open in $T_2$.

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

$\blacksquare$

Sources

• 2000: James R. Munkres: Topology (2nd ed.): $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology