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.]