Inclusion Mapping is Continuous

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $T_H = \left({H, \tau_H}\right)$ 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 $\left({\tau_H, \tau}\right)$-continuous mapping.

Proof
Let $U \in \tau$.

Then from Preimage of Subset under Inclusion Mapping:
 * $i_H^{-1} \left({U}\right) = U \cap H$

From the definition of the subspace topology:
 * $U \cap H \in \tau_H$

Hence the result by definition of a continuous mapping.