# Composition of Identification Mappings is Identification Mapping

## 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.

$\blacksquare$