I have a question about the proof of Lemma 78.12.1 from [Stacks Project](https://stacks.math.columbia.edu/tag/076L). 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](https://stacks.math.columbia.edu/tag/00WN) 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.
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> 
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Why $R=U \times\_{U/R} U$ imply that the map is injective?



Algebraic spaces in the étale topology (proof from Stacks project)
Recall from the beginning of the [proof of the lemma](https://stacks.math.columbia.edu/tag/076L) 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.


