Continuity of Composite with Identification Mapping
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Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f_1: T_1 \to T_2$ be a mapping.
Let $S_3$ be a set.
Let $p: S_1 \to S_3$ be a mapping.
Let $\tau_3$ be the identification topology on $S_3$ with respect to $p$ and $T_1$.
Let $T_3 = \struct {S_3, \tau_3}$ be the resulting topological space.
Let $f_2: S_3 \to S_2$ be a mapping such that:
- $f_1 = f_2 \circ p$
where $f_2 \circ p$ is the composition of $f_2$ with $p$.
Then $f_1$ is a continuous mapping if and only if $f_2$ is a continuous mapping.
Proof
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3.8$: Quotient spaces: Proposition $3.8.2$