Composition of Identification Mappings is Identification Mapping
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Theorem
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ and $S_3$ be sets.
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f_1$ and $\tau_1$.
Let $\tau_3$ be the identification topology on $S_3$ with respect to $f_2$ and $\tau_2$.
Let $\phi: S_1 \to S_3$ be the composition of $f_2$ with $f_1$:
- $\phi = f_2 \circ f_1$
Then $\phi$ is itself an identification mapping.
Proof
Suppose $V \subseteq S_3$ is an arbitrary subset.
We have the following chain of equivalences:
\(\ds V\) | \(\in\) | \(\ds \tau_3\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds f_2^{-1} \sqbrk V\) | \(\in\) | \(\ds \tau_2\) | $f_2$ is an identification mapping | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds f_1^{-1} \sqbrk {f_2^{-1} \sqbrk V}\) | \(\in\) | \(\ds \tau_1\) | $f_1$ is an identification mapping | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \phi^{-1} \sqbrk V\) | \(\in\) | \(\ds \tau_1\) | Preimage of Subset under Composite Mapping |
Thus, $\phi$ is an identification mapping.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 41$