The problem is: Let $(X,\mathcal{T})$ be a topological space. Suppose that given any topological space $(Y, \mathcal{F})$, any function $f: (X, \mathcal{T})\to (Y, \mathcal{F})$ is continuous. Then $\mathcal{T}$ is the discrete topology.
My answer is: In particular, let $Y=X$ and let $\mathcal{F}$ be the discrete topology and $f$ be the identity map (of sets) $X \to X$. Since $f$ is continuous, $\mathcal{T}$ must be the discrete topology.
This seems too trivial to be a correct solution. Am I missing something??
Best Answer
Your answer is perfectly fine, but I’ll show a stronger result: if $(X,\mathcal{T})$ is a space so that any map from $(X,\mathcal{T})$ to $S$ is continuous, then $X$ is the discrete space. Here $S = (\{0,1\},\{\emptyset, \{0,1\},\{0\}\})$ is the so-called Sierpiński space. So we don’t need the full strength of allowing all codomain spaces but just this fixed one.
To see this for each $x \in X$ consider the map $f_x: X \to S$ defined by $f(x)=0$, $f(x’)=1, x’ \neq x$. This is by assumption continuous and so $\{x\} = f_x^{-1}[\{0\}]$ is open in $\mathcal{T}$; as this holds for all singletons, $X$ is indeed discrete.