Injective Quotient Mapping Equals Homeomorphism

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

Let $f: S_1 \to S_2$ be a mapping.

Then $f$ is an injective quotient mapping, $f$ is a homeomorphism.

Sufficient condition
Suppose $f$ is an injective quotient mapping.

By definition of quotient mapping, $f$ is surjective.

Mapping is Injection and Surjection iff Inverse is Mapping shows that $f$ has an inverse $f^{-1}$.

To show continuity of $f^{-1}$, let $U \subseteq S_1$ be open in $T_1$.

As $f$ is bijective, it follows that $f^{-1} \sqbrk {f \sqbrk U} = U$.

By definition of quotient mapping, it follows that $f \sqbrk U$ is open in $T_2$.

As $f$ is bijective, it follows that the preimage of $U$ under $f^{-1}$ is equal to $f \sqbrk U$.

It follows that $f^{-1}$ is continuous.

By definition of homeomorphism, it follows that $f$ is a homeomorphism.

Necessary condition
Suppose $f$ is a homeomorphism.

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

Let $V \subseteq S_2$ such that $f^{-1} \sqbrk V$ is open in $T_1$.

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

As $f$ is bijective, it follows that the preimage of $f^{-1} \sqbrk V$ under $f^{-1}$ is equal to $V$.

By definition of quotient mapping, $f$ is a quotient mapping.