Final Topology Generated by Inclusion Map

Question

Let $Y\subset X$ (strict inclusion) and let $\tau^Y$ be a topology on $Y$. Then, we may endow $X$ with the final topology induced by the inclusion (map) $i:Y\rightarrow X$.

• Question: Is there an intuition behind this topology? Does it agree with the subspace topology when restricting back to $Y$?
• A Concrete Example: To help with ideas. Say $Y=C_0(\mathbb{R},\mathbb{R})$ and $X=C(\mathbb{R},\mathbb{R})$ and $Y$ is equipped with the topology of uniform convergence. What would the final topology on $X$ be? It can't be the compact-convergence topology since otherwise $\overline{Y}=X$ which is clearly not the case....

Answer 1

Well, very explicitly, a subset $U\subseteq X$ is open in the final topology iff $U\cap Y$ is in $\tau^Y$. In particular, this means every subset of $X\setminus Y$ is open, and a subset of $Y$ is open iff it is in $\tau^Y$. So, $X$ just has the topology of the disjoint union $(X\setminus Y)\sqcup Y$ where $X\setminus Y$ has the discrete topology and $Y$ has the topology $\tau^Y$. (So yes, the subspace topology on $Y$ is $\tau^Y$.)