Continuity of Composite with Inclusion

Theorem
Let $$T = \left({A, \vartheta}\right)$$ and $$T' = \left({A', \vartheta'}\right)$$ be topological spaces.

Let $$T_H = \left({H, \vartheta_H}\right)$$ be a topological subspace of $$T$$.

Let $$i: H \to A$$ be the inclusion mapping.

Let $$f: A \to A'$$ and $$g: A' \to H$$ be mappings.

Then:
 * 1) $$f \restriction_H = f \circ i$$, where $$f \restriction_H$$ is the restriction of $$f$$ to $$H$$;
 * 2) If $$f$$ is $\left({\vartheta, \vartheta'}\right)$-continuous, then so is $$f \circ i$$;
 * 3) $$g$$ is $\left({\vartheta', \vartheta_H}\right)$-continuous iff $$i \circ g$$ is $\left({\vartheta', \vartheta}\right)$-continuous;
 * 4) The induced topology $$\vartheta_H$$ is the only topology on $$H$$ satisfying (3) for all possible $$g$$.

Proof

 * 1: $$f \restriction_H = f \circ i$$:

Follows directly from the definition of the inclusion mapping.


 * 2: If $$f$$ is $\left({\vartheta, \vartheta'}\right)$-continuous, then so is $$f \circ i$$:

By Continuity of Composite Mapping, it is enough to prove that $$i$$ is $\left({\vartheta_H, \vartheta}\right)$-continuous.

This follows, because $$U \in \vartheta \Longrightarrow i^{-1} \left({U}\right) = U \cap H \in \vartheta_H$$.


 * 3: $$g$$ is $\left({\vartheta', \vartheta_H}\right)$-continuous iff $$i \circ g$$ is $\left({\vartheta', \vartheta}\right)$-continuous:

Suppose $$g$$ is continuous. Then, using the above, so is $$i \circ g$$.

Suppose then that $$i \circ g$$ is continuous.

Let $$V \in \vartheta_H$$.

Then from the definition of topological subspace, $$V = U \cap H$$ for some $$U \in \vartheta$$ and $$U \cap H = i^{-1} \left({U}\right)$$.

So $$g^{-1} \left({V}\right) = g^{-1} \left({i^{-1} \left({U}\right)}\right) = \left({i \circ g}\right)^{-1} \left({U}\right)$$.

Thus $$g^{-1} \left({V}\right) \in \vartheta'$$ by continuity of $$i \circ g$$.

Hence $$g$$ is continuous.


 * 4: Uniqueness of $$\vartheta_H$$:

Suppose $$\vartheta''$$ is a topology on $$H$$ such that:
 * For any topological space $$T' = \left({A', \vartheta'}\right)$$, and
 * For any mapping $$g: A' \to H$$,

$$g$$ is $\left({\vartheta', \vartheta''}\right)$-continuous iff $$i \circ g$$ is $\left({\vartheta', \vartheta}\right)$-continuous.

We are going to show that $$\vartheta''$$ must be the same as $$\vartheta_H$$.

First, take $$A' = H$$ and $$\vartheta' = \vartheta''$$, and $$g$$ the identity mapping of $$H$$.

Since $$g$$ is certainly $\left({\vartheta, \vartheta}\right)$-continuous, $$i \circ g$$ is $\left({\vartheta'', \vartheta}\right)$-continuous.

Hence for any $$U \in \vartheta$$, $$\left({i \circ g}\right)^{-1} \left({U}\right) \in \vartheta''$$.

But $$\left({i \circ g}\right)^{-1} \left({U}\right) = i^{-1} \left({U}\right) = U \cap H$$.

Hence $$\vartheta_H \subseteq \vartheta''$$.

Next, take take $$A' = H$$ and $$\vartheta' = \vartheta_H$$, and $$g$$ the identity mapping of $$H$$.

Since $$i \circ g = i$$ is $\left({\vartheta_H, \vartheta}\right)$-continuous, $$g$$ is $\left({\vartheta_H, \vartheta''}\right)$-continuous.

But by definition of continuity, this is the same as saying $$\vartheta'' \subseteq \vartheta_H$$.

So $$\vartheta'' = \vartheta_H$$, as required.