Smallest and largest topologies on $X$ for which $f$ is continuous This is a similar question to LINK.

Let $X$ and $Y$ be sets, and $f:X\to Y$ be a function. For a given
   topology $\mathcal{T}_{Y}$ on $Y$, describe as well as you can the
  smallest and largest topologies on $X$ for which $f$ is continuous.

Let $\mathcal{T}_{X}$ be a topology on $X$. The function $f$ is continuous iff $f^{-1}(V)\in \mathcal{T}_{X}$ for all $V\in \mathcal{T}_{Y}$. Since $f^{-1}(Y)\subseteq X$, I think I can define $\mathcal{A}_{1}=\{\emptyset, f^{-1}(Y)\}$, which is a topology on $X$. So letting $\mathcal{T}_{X}=\mathcal{A}_{1}$, the function $f$ is continuous and $\mathcal{A}_{1}$ a smallest topology on $X$. Am I on the right track?
 A: This $A_1$ defined in the question is too small. Just making $f^{-1}(Y)$ open is not enough to make $f$ continuous. Instead, in order for $f$ to be continuous, the topology should not only contain $f^{-1}(Y)$ but all $f^{-1}(V)$ for each V open in Y.
In fact, the smallest topology for $f$ to be continuous is the quotient topology induced by $f$. The definition of quotient topology induced by $f$ is that a set U in X is open if and only if $U=f^{-1}(V)$ for some V open in Y. In other words, there is no other open set in X except those which are necessary to make $f$ continuous. 
The largest topology on $X$, as is pointed out in the comment by @Alan, is the discrete topology. This is the largest possible topology on X and $f$ is definitely continuous under that topology.
A: For $f$ to be continuous we require $f^{-1}(V)$ to be open for all $V\in\mathcal T_Y$. 
Therefore, the collection $\{f^{-1}(U): U\in\mathcal T_Y\}$ must be contained in any topology on $X$. Moreover, if we can show that $\mathcal T_X=\{f^{-1}(U): U\in\mathcal T_Y\}$ is a topology on $X$ then it must be the smallest.
To show $\mathcal T_X$ is a topology we must show that it contains $X$ and $\emptyset$. We must also show that it is closed under arbitrary unions and finite intersections.
$Y\in\mathcal T_Y$ and $f^{-1}(Y)=X\in\mathcal T_X$
$\emptyset\in\mathcal T_Y$ and $f^{-1}(\emptyset)=\emptyset\in\mathcal T_X$
If $\{f^{-1(}U_i)\}_{i\in I}$ is a collection of open sets in $\mathcal T_X$ then $\cup_{i\in I}f^{-1}(U_i)=f^{-1}(\cup_{i\in I}U_i)\in\mathcal T_X$ because $\cup_{i\in I}U_i\in\mathcal T_Y$.
Finally, if $f^{-1}(U)\in\mathcal T_X$ and $f^{-1}(V)\in\mathcal T_Y$ then $f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)\in\mathcal T_X$ because $U\cap V\in\mathcal T_Y$.
