Subspace Topology on Initial Topology is Initial Topology on Restrictions

Theorem

Let $X$ be a set.

Let $I$ be an indexing set.

Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$.

Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.

Let $\tau$ be the initial topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.

Let $\struct {Y, \tau_Y}$ be a topological subspace of $X$.

Then $\tau_Y$ is the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.

Proof

From the definition of the initial topology, $\tau$ has sub-basis:

$\set {f_i^{-1} \sqbrk U : i \in I, \, U \in \tau_i}$

From Sub-Basis for Topological Subspace, $\tau_Y$ has sub-basis:

$\BB = \set {f_i^{-1} \sqbrk U \cap Y : i \in I, \, U \in \tau_i}$

That is:

$\BB = \set {\paren {f_i \restriction_Y}^{-1} \sqbrk U : i \in I, \, U \in \tau_i}$

Then $\BB$ is a sub-basis for the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.

So $\tau_Y$ is the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.

$\blacksquare$