Inclusion Mapping is Continuous/Proof 2

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$ where $H \subseteq S$.

Let $i_H: H \to S$ be the inclusion mapping on $H$.

Then $i_H$ is a $\struct {\tau_H, \tau}$-continuous mapping.

Proof

Let $1_H$ be the identity mapping on $H$.

From Continuity of Composite with Inclusion: Inclusion on Mapping, $1_H$ is $\tuple {\tau_H , \tau_H}$-continuous if and only if $i_H \circ 1_H$ is $\tuple {\tau_H , \tau}$-continuous.

From Identity Mapping is Right Identity, it follows that:

$i_H \circ 1_H = i_H$

Therefore, $1_H$ is $\tuple {\tau_H , \tau_H}$-continuous if and only if $i_H$ is $\tuple {\tau_H , \tau}$-continuous.

Because of Identity Mapping is Continuous, it follows that $i_H$ is $\tuple {\tau_H , \tau}$-continuous.

$\blacksquare$

Sources

• 2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 3$: New Spaces From Old: Subspaces