Final Topologies

# Final Topologies

Recall from the Initial Topologies page that if $X$ is a set, $Y$ is a topological space, and $f : X \to Y$ then the initial topology induced by $f$ on $X$ is the coarsest topology which makes the map $f : X \to Y$ continuous.

More generally, if $X$ is a set, $\{ Y_i : i \in I \}$ is a collection of topological spaces, and $\{ f_i : X \to Y_i : i \in I \}$ is a collection of topological maps then the initial topology induced by $\{ f_i : i \in I \}$ on $X$ is the topology which makes $f_i : X \to Y_i$ continuous for all $i \in I$.

We proved that for any set $X$ with any collection of topological spaces $\{ (Y, \tau_i) \}$ and any collection of maps $\{ f_i : X \to Y_i : i \in I \}$ that the initial topology induced by $\{ f_i : i \in I \}$ has the following subbasis:

(1)
\begin{align} \quad S = \{ f^{-1}_i(U) : U \in \tau_i, i \in I \} \end{align}

We will now look at an analogous topology known as a final topology which we define below.

 Definition: Let $X$ be a set, $\{ Y_i : i \in I \}$ be a collection of topological spaces, and $\{ f_i : Y_i \to X : i \in I \}$ be a collection of maps. The Final Topology Induced by $\{ f_i : i \in I \}$ on $X$ is the finest topology $\tau$ on $X$ which makes $f_i : Y_i \to X$ continuous for all $i \in I$.

It is important to emphasize that the final topology induced by $\{ f_i : i \in I \}$ is the FINEST topology on $X$ which makes $f_i : Y_i \to X$ continuous for all $i \in I$.

To construct the final topology on $X$ induced by the maps $f_i : Y_i \to X$, consider any of the subsets $U$ in $X$. Take the inverse image of $U$ with respect to each of the maps $f_i$, i.e., $f_i^{-1}(U)$. If $f_i^{-1}(U)$ is open in $Y_i$ for each $i \in I$, then declare $U$ to be open in $X$.

The following theorem will provide us with an explicit form of final topology induced by $\{ f_i : i \in I \}$ on $X$.

 Theorem 1: Let $X$ be a set, $\{ (Y_i, \tau_i) : i \in I \}$ be a collection of topological spaces, and $\{ f_i : X \to Y : i \in I \}$ be a collection of maps. Then the final topology induced by $\{ f_i : i \in I \}$, call it $\tau$ is given as $\tau = \{ U \subseteq X : f_i^{-1}(U) \in \tau_i \: \mathrm{for \: all \:} i \in I \}$.
• Proof: To show that $\tau$ above is the final topology induced by $\{ f_i : i \in I \}$ on $X$ we must show that $\tau$ makes $f_i : Y_i \to X$ continuous for all $i \in I$ and that any topology $\tau'$ which also accomplishes this must be coarser than $\tau$.
• Clearly each $f_i : Y_i \to X$ is continuous with this topology since for all $U \in \tau$ we have that $f^{-1}_i(U)$ is open for all $i \in I$, so $f^{-1}_i$ is continuous for all $i \in I$.
• Now suppose that $\tau'$ is another topology that accomplishes this. If $\tau \subseteq \tau'$ then there exists a $V \in \tau'$ such that $V \not \in \tau$. So $V$ is open in $X$. But then $f^{-1}_i(V)$ cannot be open in $Y_i$ for all $i \in I$ otherwise $V \in \tau$ by how we defined $\tau$. This implies that $f^{-1}_i$ is not continuous for some $i \in I$ and so $\tau'$ does not accomplish being the final topology. Hence $\tau \not \subseteq \tau'$ and so:
(2)
\begin{align} \quad \tau' \subseteq \tau \end{align}
• So any topology $\tau'$ on $X$ which also makes $f_i : Y_i \to X$ continuous for all $i \in I$ must be coarser than $\tau$, so $\tau$ is indeed the final topology induced by $\{f_i : i \in I \}$ on $X$. $\blacksquare$