Topology: functions, continuity, etc. Let $(X,T)$ be a topological space, and let $Y$ be a set (not necessarily a topological space). Define a topology $U \in Y$ as follows:
a set $U$ is defined to be open in $Y$ provided the inverse image $f^{-1}(U)$ is open in $X$ (so that $U$ is the collection of all such sets).
Prove $U$ is a topology in $Y$ and the function $f$ becomes continuous with that topology in $Y$.
Prove if $Y$ has another topology, say $V$, for which $f$ is also continuous, then $U$ is finer than $V$. (Thus, $U$ is the finest possible topology in $Y$, for which $f$ is continuous). 
 A: This is actually pretty easy once you sort out just what you have to do; I’ll help you get it organized, but I’ll leave most of the actual proof to you.
You have a topological space $\langle X,\mathscr{T}\rangle$, a set $Y$, and a function $f:X\to Y$. You’ve defined a family $\mathscr{U}$ of subsets of $Y$ by $\mathscr{U}=\{U\subseteq Y:f^{-1}[U]\in\mathscr{U}\}$. You want to prove that $\mathscr{U}$ is a topology on $Y$, that $f$ is continuous with respect to $\mathscr{T}$ and $\mathscr{U}$, and that if $\mathscr{U}'$ is any topology on $Y$ such that $f$ is continuous with respect to $\mathscr{T}$ and $\mathscr{U}'$, then $\mathscr{U}'\subseteq\mathscr{U}$. Break the problem down into smaller pieces.
To show that $\mathscr{U}$ is a topology on $Y$, you must show three things:


*

*$\varnothing,Y\in\mathscr{U}$.  

*If $U_0,U_1\in\mathscr{U}$, then $U_0\cap U_1\in\mathscr{U}$.  

*If $\mathscr{A}\subseteq\mathscr{U}$, then $\bigcup\mathscr{A}\in\mathscr{U}$.


Each of these is very straightforward. For example, if $U_0,U_1\in\mathscr{U}$, then $f^{-1}[U_0],f^{-1}[U_1]\in\mathscr{T}$, so $f^{-1}[U_0\cap U_1]=f^{-1}[U_0]\cap f^{-1}[U_1]\in\mathscr{T}$, and therefore by definition $U_0\cap U_1\in\mathscr{U}$. I’ll leave the others to you.
To show that $f$ is continuous with respect to $\mathscr{T}$ and $\mathscr{U}$, you must show that if $U\in\mathscr{U}$, then $f^{-1}[U]\in\mathscr{T}$; this is utterly trivial.
Finally, you want to assume that $\mathscr{U}'$ is a topology on $Y$ such that $f$ is continuous with respect to $\mathscr{T}$ and $\mathscr{U}'$; this means that for each $U\in\mathscr{U}'$, $f^{-1}[U]\in\mathscr{T}$, and you should have no trouble seeing from this why $\mathscr{U}'\subseteq\mathscr{U}$.
