Inclusion Mapping is Continuous

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 1

From Topological Subspace is Topological Space, $\struct {H, \tau_H}$ is a topological space.

Let $U \in \tau$.

Then from Preimage of Subset under Inclusion Mapping:

$\map {i_H^{-1} } U = U \cap H$

From the definition of the subspace topology:

$U \cap H \in \tau_H$

Hence the result by definition of a continuous mapping.

$\blacksquare$

Proof 2

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.

$\blacksquare$