Show there exists a minimal topology that makes a given function continuous 
Let $f:X \rightarrow Y$ be a function where $Y$ is a topological space. Show that there exists a topology $\tau$ in $X$ such that:
  
  
*
  
*It makes $f$ continuous.
  
*If $\tau'$ is a different topology in $X$ that makes $f$ continuous, then every open set in $\tau$ is open in $\tau'$.

I think that $\tau=\{X, \emptyset\};$ 


*

*Let $\tau_2$ be the topology defined in $Y$. Then $f$ is continuous iff for every $O \in \tau_2$, $f^{-1}(O)\in\tau$. $f^{-1}(O)=\{x\in X : f(x) \in O\}.$ Now since $\tau_2$ is a topology, then $O$ is either $\emptyset$ of a subset of Y. So if $O=\emptyset$ then $f^{-1}\{\emptyset\} = \emptyset\in\tau$. What I can't prove is the other case. I don't see why if $O$ is a subset of $Y$ then $f^{-1}\{O\}\subset X$ is the whole $X$.

*It's obvious since $\tau=\{X, \emptyset\}$ is the less fine topology in $X$
Edit:
As it was pointed out my choice of $\tau$ was wrong. Now let's consider $\tau=\{f^{-1}(U):U\text{ open in }Y\}.$ First, let's see it is a topology:


*

*$f^{-1}(\emptyset)=\emptyset$ and $f^{-1}(Y)=X$, so $\emptyset, X \in\tau.
$

*Let $\{O_i\}_{i \in I}$ be a family of open sets in $\tau.$ Then, for each $O_i, O_i = f^{-1}(U_i), U_i\in \tau_2$, so $\bigcup_{i \in I} O_i=\bigcup_{i \in I} f^{-1}(U_i) = f^{-1}(\bigcup_{i \in I} U_i)$ and since $\tau_2$ is a topology, $\bigcup_{i \in I} U_i$ is open in $\tau_2,$ so $\{O_i\}_{i \in I}=f^{-1}(\bigcup_{i \in I} U_i)$ is open in $\tau$.

*Let $O_1,O_2 \in \tau.$ Then $O_1=f^{-1}(U_1), O_2=f^{-1}(U_2), U_1, U_2 \in \tau_2.$ So $O_1\cap O_2=f^{-1}(U_1)\cap f^{-1}(U_2)=f^{-1}(U_1\cap U_2),$ and since $\tau_2$ is a topology and $U_1,U_2$ are open in $\tau_2,$ then $U_1\cap U_2$ is open in $\tau_2,$ so $O_1\cap O_2=f^{-1}(U_1\cap U_2)$ is open in $\tau.$
Now:


*

*It's clear that given $f:X \rightarrow Y,$ $\tau=\{f^{-1}(U):U\text{ open in }Y\}$ makes $f$ continuous since, by definition, $f:(X, \tau_1) \rightarrow (Y, \tau_2)$ is continuous if and only if for each $U$ open in $\taụ_2, f^{-1}(U)$ is open in $\tau_1.$

*Let $\tau'$ be a topology on $X$. Then $\tau'$ will make $f:(X, \tau') \rightarrow (Y, \tau_2) $ continuous if and only if for every $U$ open in $\tau_2,$  $f^{-1}(U)$ is open in $\tau_1.$ This means that $\tau'$ has to contain, at least, those subsets $O_i \subset X$ such that $O_i=f^{-1}(U_i), U_i \in Y,$ and the family of these subsets, $\{O_i\}_{i \in I}$ is precisely $\tau,$ so $\tau \subset \tau'$ for every $\tau'$ that makes any $f$ continuous.
 A: The topology $\{\emptyset,X\}$ on $X$ is often called the indiscrete topology.
The function $f$ need not be continuous if you endow $X$ with the indiscrete topology. Consider, for instance, $X=\{0,1\}$ and $Y$ the reals with the usual topology.
Then the map $f\colon X\to Y$ defined by $f(0)=0$ and $f(1)=1$ is not continuous with respect to the indiscrete topology on $X$, because $f^{-1}\bigl((0,\infty)\bigr)=\{1\}$ is not open.
So your conjecture is doomed, unfortunately.

Since $f$ must be continuous with respect to $\tau$, you see that $f^{-1}(U)$ should belong to $\tau$ for every open set $U$ in $Y$. Now prove that
$$
\tau=\{f^{-1}(U):U\text{ open in }Y\}
$$
is indeed the required topology on $X$
A: The topology is $\tau=\{f^{-1}(G): G~\text{open in }Y\}$, and it is the smallest one which makes $f:X\rightarrow Y$ being continuous.
A: Topology you are considering is not correct.
So correct Topology is given by,
$\tau=${$f^{-1}(A):A\in \tau_2$}, where $\tau_2$ is the topology on $Y$.
1) By the definition of continuous function, $f$ is continuous.
2) Again by the definition of continuous function and construction of topology, $\tau$ is weaker than any other topology wrt which $f$ is continuous.
