Identification Mapping is Continuous
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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.
$\blacksquare$