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I believe this hasn't been asked on this platform, so here goes. Part (b) of Corollary (1.6.2) in Gérard Laumon and Laurent Moret-Bailly's text `Champs algébriques' states the following:

Let $S$ be a scheme. Let $X \xrightarrow{f} Y \xrightarrow{g} Z$ be two morphisms of $S$-spaces (i.e. sheaves of sets on the category of $S$-schemes endowed with the étale topology). If $g \circ f$ is representable, then $f$ is representable in each of the following cases:

  1. the natural morphism $Y \times_Z Y \to Y \times_S Y$ is representable;
  2. the diagonal morphism $Z \to Z \times_S Z$ is representable;
  3. $g$ is a monomorphism.

Its proof for case (1) proceeds as follows: Pick a $S$-scheme $U$ and $y \in Y(U)$: to show $X \times_Y U$ is representable, it suffices to consider the following cartesian diagram (reader will explicate the arrows): $\require{AMScd}$ \begin{CD} X \times_Y U @>{}>> X \times_Z U\\ @VVV @VVV\\ Y \times_Z Y @>{}>> Y \times_S Y \end{CD} The proof then deduces case (2) (resp. (3)) from case (1) as follows: Note $Y \times_Z Y \to Y \times_S Y$ is deduced by base change from the diagonal morphism $Z \to Z \times_S Z$ (resp. is an isomorphism).

I have two main concerns:

  1. The above diagram does not look cartesian to me --- I only know the diagram is certainly cartesian if we replace the bottom row by the diagonal morphism $Y \xrightarrow{\Delta_g} Y \times_Z Y$, and this other case (i.e. $\Delta_g$ is representable) is the only case in which I know the corollary to be true.
  2. Also, it seems like case (3) is true, but I don't understand the explanation: if $g$ is a monomorphism, I can only deduce that the diagonal morphism $Y \xrightarrow{\Delta_g} Y \times_Z Y$ is an isomorphism (whence representable, which implies case (3)), but I cannot see why $Y \times_Z Y \to Y \times_S Y$ is an isomorphism.

Therefore, I'm asking if case (1) is true? If not, is case (2) true? Is the proof for case (3) correct?

[Edit: I'm working on my own private translation of the text, so I would appreciate if someone can confirm/deny this for me.]

Ming Hao Quek
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    Indeed, the diagram is not a pullback (= cartesian) square. In fact, the correct pullback is $X \times_Z U$ itself, since $X \times_Z U \to Y \times_S Y$ factors through $Y \times_Z Y$ and $Y \times_Z Y \to Y \times_S Y$ is a monomorphism. Case (3) is true if $Y \to Y \times_S Y$ is representable: as you say, $Y \to Y \times_Z Y$ is an isomorphism if $Y \to Z$ is a monomorphism, so in the commutative square we may replace the bottom row by $Y \to Y \times_S Y$, in which case we do have a pullback square. – Zhen Lin Sep 25 '20 at 03:02
  • @ZhenLin Thanks for confirming my doubts! As I've mentioned, I'm guessing case (1) should be corrected to: the diagonal $\Delta_g \colon Y \to Y \times_Z Y$ is representable... but do you have any idea of what the correction for case (2) should be? – Ming Hao Quek Sep 25 '20 at 11:23
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    I think I muddled up. Case (3) is true without extra hypotheses: if $Y \to Z$ is a monomorphism then $X \times_Y U \to X \times_Z U$ is an isomorphism. As for case (1) and (2), I think in both cases we should assume $Y \to Y \times_Z Y$ is representable. – Zhen Lin Sep 25 '20 at 11:53
  • @ZhenLin Oh I think I misread your earlier comment! Thanks for the reply again. – Ming Hao Quek Sep 25 '20 at 12:44

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