# Representable open immersion of functors is a monomorphism

I have a question concering the proof of theorem 8.9 in Algebraic Geometry I (U. Görtz, T. Wedhorn). I will introduce what is needed.

Let $S$ be a scheme, for an $S$-scheme $X$, we denote $h_X=Hom(-,X)$ the functor from $Sch_S^{op}$ to $Sets$.

Definition

Let $F,G:Sch_S^{op}\rightarrow Sets$ be two functors. We say that a natural transformation $f:F\rightarrow G$ is representable if for every $S$-scheme $X$ and every natural transformation $g:h_X\rightarrow G$, the functor $F\times_{G} h_X$ is representable.

Definition

Let $F,G:Sch_S^{op}\rightarrow Sets$ be two functors. We say that a representable natural transformation $f:F\rightarrow G$ is an open immersion if for every $S$-scheme $X$ and every natural transformation $g:h_X\rightarrow G$, the second projection $F\times_G h_X \rightarrow h_X$ corresponds (by Yoneda Lemma) to an open immersion of $S$-schemes $Z\rightarrow X$, where $Z$ is a representative for $F\times_G h_X$.

In the particular case of the proof of the theorem, $F_i\rightarrow F$ is a representable open immersion of functors and the $F_i$ are representable functors for every $i$. Then, citing the book

"The morphisms $F_i\rightarrow F$ are representable open immersions and open immersions are monomorphisms. Therefore the Yoneda Lemma implies that for all $S$-schemes $T$ the maps $F_i(T)\rightarrow F(T)$ are injective."

Are they meaning that we can prove that $F_i\rightarrow F$ is a monomorphism in the category of functors $Sch_S^{op}\rightarrow Sets$? Assuming this is easy to show the last assertion, but how can one prove this fact?

Or do they just want to say that from the fact that open immersion are monomorphism in the category of $S$-schemes one can deduce that the $F_i(T)\rightarrow F(T)$ are injective? I am not able to prove this weaker result neither.

## 1 Answer

Let $p : F \to G$ be an open immersion of functors. It suffices to show that each $p_T : F (T) \to G (T)$ is injective. The Yoneda lemma says there are bijections making the diagram below commute for every $T$, $$\require{AMScd}\begin{CD} \mathrm{Hom} (h_T, F) @>{\cong}>> F (T) \\ @V{\mathrm{Hom} (h_T, p)}VV @VV{p_T}V \\ \mathrm{Hom} (h_T, G) @>>{\cong}> G (T) \end{CD}$$ but for each $x_0, x_1 : h_T \to F$, if $p \circ x_0 = p \circ x_1$, then $$\begin{CD} h_T @>{x_0}>> F \\ @| @VV{p}V \\ h_T @>>{p \circ x_1}> G \end{CD}$$ must be a pullback diagram, so there is a unique morphism $a : h_T \to h_T$ such that $x_0 \circ a = x_1$ and $a = \mathrm{id}$, hence $x_0 = x_1$. Therefore each $p_T : F (T) \to G (T)$ is injective, as claimed.

• Thank you for the answer, but I do not get why the last diagram should be a pullback diagram and where you use the hypothesis on $p$. – Pgatti Dec 1 '14 at 17:37
• Your first question is answered by the second: I use the hypothesis on $p$ to show that the diagram is a pullback square. – Zhen Lin Dec 1 '14 at 17:44
• I tried to work it out but I did not succeed. Can you, please, show me some more details on how to prove that the diagram is a pullback? Thank you. – Pgatti Dec 1 '14 at 20:45
• By hypothesis, $h_T \times_G F$ is (the representable functor associated with) an open subscheme of $T$, but $\mathrm{id}_{h_T} : h_T \to h_T$ factors through the projection $h_T \times_G F \to h_T$, so in fact this open subscheme is $T$ itself. – Zhen Lin Dec 2 '14 at 1:22