Initial Topologies

Initial Topologies

Now that we have looked at the definition of continuity of maps on topological spaces, we are now ready to look at some new topologies that are based off of the concept.

We begin by describing a special type of topology on a set $X$ that is induced by a topological space $Y$ by a function $f : X \to Y$ (or a collection topological spaces and functions $f_i : X \to Y_i$) ensuring continuity of $f$ (or each $f_i$, $i \in I$).

 Definition: Let $X$ be a set and let $Y$ be a topological space. The Initial Topology Induced by $f$ on $X$ is the coarsest topology on $X$ that makes the map $f : X \to Y$ continuous. Furthermore, if $X$ is a set and $\{ Y_i : i \in I \}$ is a collection of topological spaces then the Initial Topology Induced by $\{ f_i : i \in I \}$ on $X$ is the coarsest topology on $X$ that makes each map $f_i :X \to Y_i$ continuous.

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

So, consider a set $X$ and a topological space $Y$. Let $f : X \to Y$ and consider the set of all open sets in $Y$. Then for $f$ to be continuous we must have that the inverse image of all open sets in $Y$ are open in $X$. Thus, if we give $X$ the initial topology then the open sets of $X$ are just that - sets such that for every open set $V$ in $Y$ we have that $f^{-1}(V)$ is declared an open set in $X$. We exclude adding any additional open sets to $X$ if possible since the initial topology on $X$ induced by $f$ is the coarsest of such topologies. We will need something more than just a wordy definition if we're expecting to work with initial topologies induced by $\{ f_i : i \in I \}$, so, the following theorem will give us a subbasis for this topology.

 Theorem 1: Let $X$ be a set, $\{ (Y_i, \tau_i) : i \in I \}$ be a collection of topological spaces, and $\{ f_i : X \to Y_i : i \in I \}$ be a collection of maps. Then the initial topology induced by $\{ f_i : i \in I \}$ on $X$ has subbasis $S = \{ f^{-1}_i (U) : U \in \tau_i \}$.
• Proof: To show that the initial topology induced by $\{ f_i : i \in I \}$ on $X$ has subbasis $S = \{ f^{-1}_i(U) : U \in \tau_i \}$ we must show that the topology $\tau$ generated by this subbasis is equal to this induced initial topology on $X$. To do this, we must show that $\tau$ is the topology that makes each $f_i : X \to Y_i$ for $i \in I$ continuous, and we must show that $\tau$ is the finest topology to do this.
• Clearly the topology $\tau$ generated by subbasis $S$ makes all of the $f_i : X \to Y$, $i \in I$ continuous, because for each $U \in \tau_i$ we have that $f^{-1}(U) \in \tau$ (is open in $X$ with respect to the topology $\tau$ generated by the subbasis $S$). Hence $f_i$ is continuous for all $i \in I$.
• We now show that $\tau$ is the coarsest topology to accomplish this. Suppose that $\tau'$ is another topology that makes $f_i : X \to Y_i$ continuous for all $i \in I$. Then $f^{-1}_i(U_i) \in \tau'$ for all $i \in I$. But since $\tau'$ is a topology on $X$ we have that all finite intersections of $f^{-1}_i(U_i)$ are also contained in $\tau'$ and so:
(1)
\begin{align} \quad \tau \subseteq \tau' \end{align}
• So any such topology $\tau'$ which also accomplishes making $f_i : X \to Y_i$ continuous for all $i \in I$ must contain $\tau$, so the initial topology $\tau$ induced by $\{ f_i : i \in I \}$ on $X$ has subbasis $S = \{ f^{-1}_i(U) : U_i \in \tau_i \}$. $\blacksquare$