Quotient Mapping and Continuous Mapping Induces Continuous Mapping

Theorem
Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$, $T_3 = \struct {S_3, \tau_3}$ be topological spaces.

Let $p: S_1 \to S_2$ be a quotient mapping.

Let $g: S_2 \to S_3$ be a mapping such that for all $s_1, s_2 \in S_1$ with $\map p {s_1} = \map p {s_2}$, we have $\map g {s_1} = \map g {s_2}$.

Then $g$ induces a mapping $f: S_2 \to S_3$ such that $f \circ p = g$.

The induced mapping $f$ is a continuous mapping, $g$ is a continuous mapping.


 * $\begin{xy} \xymatrix@L+2mu@+1em{

S_1 \ar[r]^*{p} \ar[rd]_*{g} & S_2 \ar@{-->}[d]^*{f} \\ & S_3 }\end{xy}$

Proof
For all $t \in S_2$ we can find $s \in S_1$ with $\map p s = t$, as $p$ is surjective.

Define $\map f t := \map g s$.

For all $s' \in S_1$ with $\map p {s'} = t$, we have $\map g s = \map g {s'}$ by assumption, so $f$ is well-defined.

It follows that $\map g s = \map {f \circ p} s$.

Suppose $f$ is continuous.

As $p$ is continuous by definition of quotient mapping, Composite of Continuous Mappings is Continuous shows that $g = f \circ p$ is continuous.

Suppose $g$ is continuous.

Let $V \subseteq S_3$ be open in $T_3$.

As $g$ is continuous, it follows that $g^{-1} \sqbrk V$ is open is $T_1$.

As $g = p \circ f$, we have $g^{-1} \sqbrk V = p^{-1} \sqbrk { f^{-1} \sqbrk V }$.

As $p$ is a quotient mapping, we have $p \sqbrk { p^{-1} \sqbrk { f^{-1} \sqbrk V } }$ open in $T_2$.

Image of Preimage under Mapping/Corollary shows that $p \sqbrk { p^{-1} \sqbrk { f^{-1} \sqbrk V } } = f^{-1} \sqbrk V$.

As $f^{-1} \sqbrk V$ is open in $T_2$, it follows that $f$ is continuous.