# Find the coarsest topology?

Let $X$ be a set and $(Y,\tau)$ be a topological space. Let $f_i : X \longrightarrow Y, (i\in I)$ be a collection of maps. Find the COARSEST topology on $X$ such that all maps $f_i, (i\in I)$ are continuous.

• Well, determine which subsets of $X$ you absolutely need to be open, and work from there. – Tara B Mar 14 '13 at 17:35
• Don't you simply want to say that the collection $\{ f_i^{-1}[U]\mid i\in I \text{ and$U$open in$Y$}\}$ forms a subbasis for the desired topology? – MJD Mar 14 '13 at 17:39

HINT: You know that for each $i\in I$ and $U\in\tau$ the set $f_i^{-1}[U]$ must be open in $X$, so let $$\mathscr{S}=\left\{f_i^{-1}[U]:i\in I\text{ and }U\in\tau\right\}\;;$$ what is the coarsest topology $\tau'$ on $X$ such that $\mathscr{S}\subseteq\tau'$? (This topology $\tau'$ is sometimes called the initial topology with respect to the family $\{f_i:i\in I\}$ of functions.)
• Thank you. I choose $\tau' = {\bigcap_{i\in I}\left(\bigcup_{U\in\tau}f_{i}^{-1}\left[U\right]\right)}$ It true??? – RioDejaneiro Mar 15 '13 at 14:47
• @ĐặngPhướcNhật: That’s $$\tau'=\left\{\bigcap_{i\in I}\left(\bigcup_{U\in\tau}f_i^{-1}\left[U\right]\right)\right\}\;.$$ No, that makes $\tau'=\{X\}$. You want the topology generated by $\mathscr{S}$ as a subbase. Start by forming the collection of all intersections of finitely many sets of the form $f_i^{-1}[U]$ with $U\in\tau$, and then take arbitrary unions of these intersections. – Brian M. Scott Mar 15 '13 at 14:50
• $\tau'=\left\{\bigcup_{U\in \tau}\left(\bigcap_{i\in I}f_i^{-1}\left[U\right]\right)\right\}$ ?? – RioDejaneiro Mar 15 '13 at 16:46
• @ĐặngPhướcNhật: Once again you’ve made $\tau'$ a single set. You cannot write it this simply. You’d be well advised to write the definition in two steps, as I described it in my previous comment. Let $$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\text{ is a finite subset of }\mathscr{S}\right\}\;,$$ and then let $$\tau'=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;.$$ – Brian M. Scott Mar 15 '13 at 16:50
What sets do we require to be open in $X$ in order for $f_i:X\to Y$ to be continuous? (Hint: use the definition of continuity) Call the collection of those sets $C_i$. Is there at least one topology $S$ on $X$ such that $S\supseteq\bigcup_{i\in I}C_i$? (Hint: what is the finest topology on $X$?) Show that the intersection of all such topologies $S$ is again topology on $X$, and that in fact it is the desired coarsest topology.