# Algebraic spaces in the étale topology (proof from Stacks project)

I have a question about the proof of Lemma 78.12.1 from Stacks Project. The aim of the last paragraph of the proof is to verify that the map of sheaves in the étale topology $$F \to U/R$$ is an isomorphism. By Lemma 7.11.2 our job is to show that it's surjective and injective. The proof that $$F \to U/R$$ is injective I not understand. The used argument is:

On the other hand, the map $$F \to U/R$$ is injective (as a map of presheaves) since $$R=U \times_{U/R} U$$ again by Spaces, Theorem 63.10.5.

Why $$R=U \times_{U/R} U$$ imply that the map is injective?

Recall from the beginning of the proof of the lemma that $$R$$ is defined to be $$U \times_F U$$, so the surjections $$U \to F \to U/R$$ induce a canonical étale sheaf map $$U \times_{U/R} U \to U \times_F U$$ that was shown to be an isomorphism.
Let's see what this means for a test object $$S$$. Let $$A = U(S)$$, $$B = F(S)$$, $$C = (U/R)(S)$$, and let $$f: A \to B$$ and $$g: B \to C$$ be induced by the maps $$U \to F \to U/R$$ given before. Then we are basically saying that, if we replace $$S$$ with a suitably fine étale cover, then the induced set map $$A \times_C A \to A \times_B A$$ is an isomorphism. Here, $$A \times_C A = \{ (x,y) \in A \times A | gf(x) = gf(y) \}$$ and $$A \times_B A = \{ (x,y) \in A \times A | f(x) = f(y) \}$$. That is to say, $$gf(x) = gf(y)$$ if and only if $$f(x) = f(y)$$, or $$g$$ is injective on the image of $$f$$.
Because $$U \to F$$ is a surjection of sheaves, we may assume $$f$$ is surjective after passing to a suitable cover. We conclude that $$g$$ is injective, and hence $$F \to U/R$$ is injective.
• everything is clear now except one step: Why is the induced $U \times_{U/R} U \to U \times_F U$ is an isomorphism? – katalaveino Aug 14 at 16:49
• yes, I think I understand. Theorem 63.10.5 implies that since $R := U \times_F U \to U \times_S U$ is etale equiv relation, the data $(U,R, U \to U/R)$ is representation of $U/R$. But beeing represented by this data implies for $U/R$ tautologically that $R=U \times_{U/R} U$ holds? that's the point, right? – katalaveino Aug 15 at 1:43