Open Continuous Injection is Embedding
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 an open and continuous injection.
Then $f$ is an embedding of $T_1$ into $T_2$.
Proof
Let $g: S_1 \to f \sqbrk {S_1}$ be the restriction of $f$ to $S_1 \times f \sqbrk {S_1}$.
It must be shown that $g$ is a homeomorphism.
$g$ is a bijection
From Restriction of Mapping to Image is Surjection, $g$ is a surjection.
From Restriction of Injection is Injection, $g$ is a injection.
Thus, $g$ is a bijection by definition.
$\Box$
$g$ is continuous
Follows from Restriction of Continuous Mapping is Continuous: Topological Spaces
$\Box$
$g$ is open
Let $A \subseteq S_1$ be open in $T_1$.
Since $f$ is an open mapping, $f \sqbrk {A}$ is open in $T_2$.
Because $f$ is an open mapping, $f \sqbrk {S_1}$ is open in $T_2$, as $S_1$ is open in $T_1$ by Definition of a topology.
Thus Open Set in Open Subspace applies: $f \sqbrk {A}$ is open in $T_2$ if and only if it is open in the subspace topology of $f \sqbrk {S_1}$.
Therefore, $f \sqbrk {A}$ is open in the subspace topology of $f \sqbrk {S_1}$.
Since $f$ and $g$ agree on $S_1$, $f \sqbrk {A} = g \sqbrk {A}$.
Therefore, $g$ is an open mapping.
$\Box$
$\blacksquare$
Also See
Sources
- 2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 3$: New Spaces From Old: Subspaces. Topological Embeddings