Describing continuous functions from X to Y

Question

I want to describe all continuous functions from $X$ to $Y$ where $X$ is the indiscrete topology. I have proved that if $Y$ is Hausdorff then $f$ must be constant, but my supervisor wants me to do it where $Y$ is a general topology. I would appreciate any hints/ideas.

Answer 1

It is necessary and sufficient that the elements in the image are topologically indistinguishable, that is, that all of them belong to exactly the same open sets.

Indeed, if $f:X \to Y$ is so, then for a an open set $V \subseteq Y$, we have that either $f^{-1}(V) = \varnothing$, if $f(X) \cap V = \varnothing$, or $f^{-1}(V) = X$, since if $f(X) \cap V \neq \varnothing$, then $f(X) \subseteq V$.

Conversely, if there are $a,b \in X$ with $f(a)$ topologically distinct from $f(b)$, say there exists an open set $U$ such that $f(a) \in U$ and $f(b) \notin U$, then, since $a \in f^{-1}(U)$, we have $f^{-1}(U) \neq \varnothing$, whence $f^{-1}(U)=X$ and so $f(b) \in U$, a contradiction.

Answer 2

Constant functions will be continuous for any topology in $Y$.

The question is, given $X$ with the indiscrete topology, what set are you looking for:

1. $\{f: X \to (Y,\tau_Y) ; \quad \tau_Y \quad \textrm{is a topology in } Y \ \textrm{and f is continuous}\}$ or
2. Given a fixed topology in $Y$ the set of all continuous functions from $X$ to $Y$ ?

If its the first case, then your only options are constant functions, it is easy to prove.

If it is the second case, apart from the constant function, I think the only functions are:

The ones that map $X$ to an open set $U \subset Y$ given that $U$ is the smallest (in the sense of inculsion) open set contained in $U$.

Because the pre-image any other open set that does not contain $U$ is the empty set, and the pre-image of a set containing $U$ is $X$.

Or the ones that map $X$ to the intersection of all non-empty closed sets in $Y$.

Because the pre-image of any open set (excluding Y) will be the empty set, and the pre-image of $Y$ is $X$.

Any other one will not do.