I found this question: $f$ is continuous if and only if $f$ is constant
But I wondered whether all the assumptions are necessary. I think we can state the following claim as well:
Let $f:X_1\rightarrow X_2$ be a continuous map, where $X_1$ has the trivial topology and $X_2$ has a $T_1$ topology. Then $f$ is constant.
My argument is simple:
Assume towards contradicition that $y_1,y_2\in X_2$ are distinct points attained by $f$. Then $f^{-1}[\{y_1\}]$ is closed and not $X_1$, which is a contradicition. Hence $f$ must be constant.
My question is whether I can reduce the requirement on $X_2$ to simply be a $T_0$ space? I think such an argument would be true for $y_1,y_2\in X_2$ attained by $f$ such that we have a neighbourhood $U\subset X_2$ such that $y_1\in U$ and $y_2\in U$.
Best Answer
Yes, it is enough if $X_2$ is a $T_0$ space.
Then if a function is not constant, i.e its image contains at least $2$ distinct points, there is an open set $U$ containing exactly one of these points.
The preimage of $U$ cannot be an element of $\{\varnothing, X_1\}$ so the function is not continuous.