So I'm led to believe this is a fairly standard result. See, for example Lemma 25.15.4 here. I am trying to prove a criterion for the representability of a (contravariant) functor from schemes to sets. I'm having a bit of difficulty understanding a step in the proof in the link.

Six lines into the proof the sentence beginning:

"Since $(\varphi_{ij} \circ \varphi_{ji})^{*} \xi_{j} = \varphi_{ji}^{*}(\varphi^{*}_{ij} \xi_{j}) = \varphi_{ji}^{*} \xi_{i} = \xi_{j}$ we..."

I'm not entirely sure what these equalities even mean. It seems that my confusion is coming from the fact that they are freely identifying the morphisms $\varphi_{ij}: U_{ij} \rightarrow X_{j}$ with the morphism via which it factors through $U_{ji}$. In the above sequence of equalities, in the notation of the linked proof, say we have the subfunctor $F_{j} \subset F$ so that, $$ (\varphi_{ij} \circ \varphi_{ji})^{*} \xi_{j} $$ is given by the image of $\xi_{j}$ under the morphism, $$ F_{j}(X_{j}) \stackrel{F_{ij}(\varphi_{ij})}{\longrightarrow} F_{j}(U_{ij}) \subseteq F_{j}(X_{j}) \stackrel{F_{j}(\varphi_{ji})}{\longrightarrow} F_{j}(U_{ji}) $$ Is this the correct interpretation? I think(??) the subset arises from the fact that the representable functor takes an open immersion to an injective function of sets.

But then how do I interpret the image of that composition, which is an element in $F_{j}(U_{ji})$, as the element $\varphi^{*}_{ji} \xi_{i}$, which I think should be an element in $F_{i}(U_{ji})$? Do we interpret both as subsets of $F(U_{ji})$?

I'm hoping I've made it clear where exactly my confusion lies, although I'm not sure I know myself. Is someone able to elaborate a bit on what elements are in what set, and how the sequence of equalities above is interpreted?


The formulas in the proof you linked are not literally true.

Since $\varphi_{ij}$ factors through $U_{ji}$, we can replace it by the induced morphism $U_{ij}\to U_{ji}$. Analogously for $\varphi_{ji}$. Then the composite $\varphi_{ij}\circ\varphi_{ji}$ makes sense, and we compute

$$(\varphi_{ij}\circ\varphi_{ji})^*(\xi_j|_{U_{ji}}) = \varphi_{ji}^*(\varphi_{ij}^*(\xi_j|_{U_{ji}})) = \varphi_{ji}^*\xi_i|_{U_{ij}} = \xi_j|_{U_{ji}}.$$

Now conclude that $$ U_{ji}\xrightarrow{\varphi_{ji}}U_{ij}\xrightarrow{\varphi_{ji}}U_{ji}\to X_j $$ must be equal to $U_{ji}\to X_j$, and use that $U_{ji}\to X_j$ is monic to deduce $\varphi_{ij}\circ\varphi_{ji}=\mathrm{id}_{U_{ji}}$.

[NB. What's really happening here is the following: $U_{ij}$ represents $F_i\times_FF_j$, and $U_{ji}$ represents $F_j\times_FF_i$; since these functors are isomorphic, we find a unique isomorphism between $U_{ij}$ and $U_{ji}$ making the universal elements correspond.]

  • $\begingroup$ Thank you for the answer, I appreciate it. That definitely takes care of the first equality, but I am still unsure of the other two. Take the element $\phi_{ji}^{*} \xi_{i}|_{U_{ij}}$ for instance. Presumably the term $\xi_{i}|_{U_{ij}}$ refers to the image of $\xi_{i}$ under the function $F_{i}(X_{i}) \rightarrow F_{i}(U_{ij})$ induced by the inclusion $U_{ij} \hookrightarrow X_{i}$. This lands you in $F_{i}(U_{ij})$, but the function $\phi^{*}_{ji}$ maps from $F_{j}(U_{ij})$. $\endgroup$ – Luke May 20 '18 at 15:37
  • $\begingroup$ @Luke: The way I argued above (checking that two morphisms to $X_j$ agree), the stars and bars stand for the action of $F_j$. Note that, by definition, $\xi_i|_{U_{ij}}\in F_j(U_{ij})$. $\endgroup$ – user501746 May 21 '18 at 9:55

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