Identification Mapping is Continuous

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.

Let $S_2$ be a set.

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

Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.

Then the identification mapping $f$ is continuous.

Proof
By definition of identification topology:
 * $\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$

That is:
 * $V \in \tau_2 \implies f^{-1} \sqbrk V \in \tau_1$

This is precisely the definition of a continuous mapping.