Let $\mathfrak T_X = \{f^{-1} (U) : U \in \mathfrak T_Y\}$ then $\mathfrak T_X$ is a topology on X. False? Let $f :X \rightarrow Y$ be a function and suppose that $\mathfrak T_Y$ is a topology on $Y$.  Let $\mathfrak T_X = \{f^{-1} (U) : U \in \mathfrak T_Y\}$ then $\mathfrak T_X$ is a topology on X. 
$f^{-1}(U)$ refers to the inverse image of a set
I am supposed to determine if this is true or false and then prove or give a counterexample. I am leaning towards this being a false statement because this conjecture doesn't say anything about the function being continuous but I am not sure if this is necessary.  Can anyone help with where to go with my thinking?
 A: First ask yourself what does it mean for function $f:X\to Y$ to be continuous? It is defined that for every open subset $U$ of $Y$ we will have that $f^{-1}(U)$ is a open subset of $X$. But if we didn't defined yet a topology on $X$ how can we even refer to the continuity of $f$?
In the claim, what comes in to question is whether $\mathfrak T_X$ defines a topology on $X$ and not whether $f$ is continuous. note that if it is a topology then indeed $f$ will be continuous. Another fact is, that $\mathfrak T_X$ would have been the most coarser (or least finer) topology on $X$ such that $f$ would have been continues. 
so try to see weather $\mathfrak T_X$ satisfy the definition of a topology. and say if you need more help.

We would like to show the following: 
1) $X,\emptyset\in\mathfrak T_X$
2) if $\{U_a\}_{a\in I}\subseteq\mathfrak T_X$ for some idex set $I$, then $\cup _{a\in I}U_a$ 
3) if $\{U_i\}_{i=1}^n\subseteq\mathfrak T_X$ then $\cap_{i=1}^nU_i\in\mathfrak T_X$ 
For (1) it is immediate that $f^{-1}(Y)=X$ and $f^{-1}(\emptyset)=\emptyset$.
For (2) and (3) you will want to prove two identities about functions.
a) $f^{-1}(\cup _{a\in I}U_a)=\cup_{a\in I} f^{-1}(U_a)$ 
b)  $f^{-1}(\cap _{i=1}^nU_i)=\cap_{i=1}^n f^{-1}(U_i)$ 
and from here I hope that the rest of the proof is clear to you.
Note that the collection of $U_a$'s need not be in $\mathfrak T_X$ for those two identities to hold. 
