Definition 3.1.8 (Final Topology).label Let $X$ be a set, $\curl{(Y_i,\tau_i)}_{i\in I}$ be a family of topological spaces, and $\curl{f_i:Y_i\to X}_{i\in I}$ be a family of maps then there exists a topology $\tau$ on $X$ satisfying:
- (1)
$f_{i}\in C(Y_{i};X)$ for each $i\in I$
- (U)
$\tau$ is the finest topology on $X$ such that all $f_{i}$ are continuous
- (U’)
For any topological space $Z$ and map $g:X\to Z$ we have $g$ is continuous if and only if $g\circ f_{i}:Y_{i}\to Z$ is continuous for each $i\in I$.
- (4)
(1)+(U)$\iff$(U’)
called the final/strong topology on $X$ generated by the maps $\curl{f_i}_{i\in I}$.
Proof. Let $\tau\define \curl{U\suf X:\fall i\in I,f_i\inv(U)\in\tau_i}$ then $\tau$ is a topology since preimages commute with union and intersection. Furthermore
- (1)
By definition $U\in\tau\implies f_{i}\inv(U)\in\tau_{i}$ for each $i\in I$ hence $f_{i}\in C(Y_{i};(X,\tau))$ for each $i\in I$
- (U)
If $\tau'$ is a topology on $X$ satisfying (1) then for each $U\in\tau'$, $f_{i}\inv(U)\in\tau_{i}$ for all $i\in I$ so $U\in \tau$ so $\tau'\suf \tau$.
- (U’)
If $g$ is continuous then $g\circ f_{i}$ is continuous for each $i\in I$ since composition of continuous functions is continuous. Let $i\in I$ and $V\suf Z$ open then
\begin{align*}f_{i}\inv(g\inv(V))=(g\circ f_{i})\inv(V)\end{align*}is open in $Y_{i}$ by continuity of $g\circ f_{i}$ as desired.
- (4)
Clear
$\square$
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