Continuous Surjection Induces Continuous Bijection from Quotient Space

Theorem
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.

Let $g: S_1 \to S_2$ be a continuous surjection.

Let $\RR_g \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $g$:
 * $\tuple {s_1, s_2} \in \RR_g \iff \map g {s_1} = \map g {s_2}$

Let $q_{\RR_g}: S_1 \to S_1 / \RR_g$ be the canonical surjection induced by the equivalence relation $\RR_g$.

Let $\tau_{\RR_g}$ be the quotient topology on the quotient space $S_1 / \RR_g$ by $q_{\RR_g}$.

Then $g$ induces a continuous bijection $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.

Corollary 2

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

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

Proof
By definition of quotient topology, $q_{\RR_g}$ is a surjective identification mapping.

Quotient Mapping equals Surjective Identification Mapping shows that $q_{\RR_g}$ is a quotient mapping.

Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $g$ induces a continuous mapping $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.

As $g$ is surjective, it follows that $f$ is surjective.

Let $\eqclass {s_1}{\RR_g}, \eqclass {s_2}{\RR_g} \in S_1 / \RR_g$ such that $\map f { \eqclass {s_1}{\RR_g} } = \map f { \eqclass {s_2}{\RR_g} }$.

As $f \circ q_{\RR_g} = g$, it follows that $\map g {s_1} = \map g {s_2}$.

By definition of equivalence relation induced by $g$, it follows that $\eqclass {s_1}{\RR_g} = \eqclass {s_2}{\RR_g}$.

By definition of injection, it follows that $f$ is injective.

As $f$ is injective and surjective, it follows that $f$ is bijective.