I'm trying to understand the construction of the locus of agreement of schemes. Multiple references use a certain commutative diagram but I fail to see why it's commutative.

Let $ f,g:X\rightarrow Y$ be two $ S $-scheme morphisms. The locus of agreement of $ f $ and $ g $ is defined to be $V$, the pullback of the diagram $$\require{AMScd}\begin{CD}V@>>>Y\\@VVV@V\Delta VV\\X@>f \times g>>Y\times_SY\end{CD}$$

Let $h:Z\rightarrow X$ be an $ S $-morphism so that $ f\circ h=g\circ h$. Then $ h $ factors uniquely through $ V $.

Why does $ Z $ form a commutative diagram with $ \Delta $ [the diagonal] and $ h $ [unique factorization through the pullback]?


  • 1
    $\begingroup$ I think you have an error in your diagram: the lower right corner is currently $X\times_S Y$, but should be $Y\times_S Y$. $\endgroup$
    – KReiser
    Dec 30 '14 at 21:37
  • $\begingroup$ @KReiser Right. I corrected that. $\endgroup$
    – PeterM
    Dec 30 '14 at 21:57

If $h: Z\to X$ is an $S$-morphism so that $f\circ h = g\circ h$, then $f\circ h$ and $g\circ h$ are identical maps from $Z\to Y$. In addition, we have that $(f\times g)\circ h$ and $\Delta \circ (f\circ h) = \Delta \circ (g\circ h)$ are equal in $Y\times_S Y$. So $h:Z\to X$ and $f\circ h=g\circ h : Z\to Y$ are maps from $Z$ to $X,Y$ such that the composites agree in $Y\times_S Y$. So by the universal property of pullbacks, we have a unique morphism $Z\to V$.


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