Quotient Mapping equals Surjective Identification 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 mapping.

Then $f$ is a quotient mapping, :


 * $f$ is surjective, and $\tau_2$ is the identification topology on $S_2$ with respect to $f$ and $T_1$.

Sufficient condition
Let $f$ be surjective, and $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $T_1$.

Identification Mapping is Continuous shows that $f$ is continuous.

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

By definition of identification topology, it follows that $U$ is open in $T_2$.

Hence $f$ fulfills all the requirements to be a quotient mapping.

Necessary condition
Suppose $f$ is not surjective, or $\tau_2$ is not the identification topology on $S_2$ with respect to $f$ and $T_1$.

If $f$ is not surjective, then $f$ is by definition not a quotient mapping.

Existence and Uniqueness of Identification Topology shows that the identification topology is uniquely defined.

So if $\tau_2$ is not the identification topology, there exists a subset $U \subseteq S_2$ not open in $T_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.

Then $f$ does not fulfill the condition to be a quotient mapping.