My question is similar to the question in The image of the diagonal map in scheme. I saw a hint to my question in the comments, but I was not able to prove it.

Let $f:X\longrightarrow Y$ is a morphism of schemes. Let $T$ be a scheme and $h:T\longrightarrow X\times_Y X$ be a morphism such that $p_1\circ h=p_2\circ h$, where $p_1,p_2$ are projection morphisms from $X\times_Y X\longrightarrow X$. Then is it true that $h(T)\subset\Delta(X)$?

I initially tried proving that if $x\in X\times_Y X$ such that $p_1(x)=p_2(x)$, then $x\in\Delta(X)$. But that I think is not true, and I am going nowhere. I will be help if someone can tell me how to prove it.


Don't work with elements, work with morphisms. This general lesson of category theory is also very useful for algebraic geometry.

Let $g := p_1 h = p_2 h : T \to X$. Then $h$ is equal to $T \xrightarrow{g} X \xrightarrow{\Delta} X \times_Y X$, since $p_i \Delta g = g = p_i h$. Hence, $h$ factors through $\Delta$.

By the way, you are right with your guess that $z \in X \times_Y X$ with $x:=p_1(z)=p_2(z)$ doesn't imply that $z \in \Delta(X)$. We also need that $(p_1^\#)_z : \mathcal{O}_{X,x} \to \mathcal{O}_{X \times_Y X,z} $ is equal to $(p_2^\#)_z : \mathcal{O}_{X,x} \to \mathcal{O}_{X \times_Y X,z}$. For example, let $X=\mathrm{Spec}(L)$ and $Y=\mathrm{Spec}(K)$ for a field extension $L/K$. Then $p_1(z)=p_2(z)$ is a vacuous condition! The image of $\Delta : \mathrm{Spec}(L) \to \mathrm{Spec}(L \otimes_K L)$ has just one element corresponding to the kernel of the multiplication $L \otimes_K L \to L$, but $L \otimes_K L$ usually has more prime ideals (if $L$ is simple over $K$, say $L=K[x]/(f)$, then $L \otimes_K L = L[x]/(f)$ and the prime ideals of $L \otimes_K L$ correspond to the irreducible factors of $f$ over $L$).

  • $\begingroup$ Thanks a lot! Things have become clear! $\endgroup$ – gradstudent Dec 16 '13 at 16:32
  • $\begingroup$ Yes I just remembered that, and so deleted the comment :) $\endgroup$ – gradstudent Dec 16 '13 at 16:33
  • $\begingroup$ If $ p_1\circ h$ and $ p_2\circ h $ agree at a point $ t $, is it still true that $ h (t) \in \Delta (X) $. $\endgroup$ – gradstudent Dec 16 '13 at 18:01
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    $\begingroup$ This statement is not true in general. It holds when $p_1 h = p_2 h$ as morphisms. $\endgroup$ – Martin Brandenburg Dec 16 '13 at 18:27

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