What is a constant morphism of schemes

Question

I am dealing with some basic notions on schemes and I was asking myself what does it mean for a morphism between two schemes $$f:(X,\mathcal{O}_X)\rightarrow(Y,\mathcal{O}_Y)$$ to be constant. The obvious answer is that $f(X)$ is reduced to a point $\{y\}$, however I was wondering what happens on the level of the sheaf morphism: $$f^\#:\mathcal{O}_Y\rightarrow f_*\mathcal{O}_X$$ Now for any open $U\subseteq Y$, one has $$f_*\mathcal{O}_X(U)=\left\{\begin{array}[c], X\text{ if }y\in U\\ \emptyset\text{ if }y\notin U \end{array}\right.$$ thus ignoring the canonically zero maps $\mathcal{O}_Y(U)\rightarrow f_*\mathcal{O}_X(U)=0$, when $y\notin U$, this amounts to a system of maps $\big(\mathcal{O}_Y(U)\rightarrow f_*\mathcal{O}_X(U)=\mathcal{O}_X(X)\big)_{y\in U}$ compatible with restriction, ie a morphism $$\mathcal{O}_{Y,y}\rightarrow\mathcal{O}_X(X)$$ thus à priori there are more than just one constant morphism with image $\{y\}$. Am I missing something? If not, are there some classic examples that can illustrate this fact? Any input is appreciated.

Answer 1

That’s not the right definition. In any category with a terminal object $1$, a constant morphism is a morphism that factors through the terminal object. In particular there is exactly one such morphism $X \to Y$ for every global point $1 \to Y$; note that these are very different from points of the Zariski spectrum, for schemes, because the terminal object is $\text{Spec } \mathbb{Z}$, and in particular there often won’t be any. E.g. with this definition there are no constant morphisms if $Y$ is the spectrum of a field.

The nLab also suggests another definition which is slightly weaker.