The Stacks project

I don't see how bijection is proved in the general scheme case. It's not that the proof is wrong, but morelike it's not enough. The rest of the proof still needs some non-trivial logic. $\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Spec}{Spec}$ Define $$c:\Hom(\Spec R,X)\to \\{ (x,\varphi),x\in X,\varphi\in \Hom_{local}(\mathcal{O}\_{X, x},R)\\},f\mapsto (f(\mathfrak{m}),f^\sharp)$$ as above. We first show it's surjective. Given any $(x,\varphi)$, there exists an open affine $U=\Spec A$ containing $x$, and a map $A \to \mathcal{O}\_{X, x}$, by composing with $\varphi:\mathcal{O}\_{X, x}\to R$, we have $f:\Spec R\to \Spec A\to X$. It's easy to see $c(f)=(x,\varphi)$. Next we show it's injective. Given any two maps $g,h:\Spec R\to X$, s.t. $c(g)=c(h)=(x,\varphi)$. We know $g(\mathfrak{m})=h(\mathfrak{m})=x$. Pick an open affine $U=\Spec A$ containing $x$ we know both $g$ and $h$ factor through $U$ using the orginal argument. So we have $g_0,h_0:\Spec R\to U$. It suffices to show $g_0=h_0$. Clearly $c$ is functorial w.r.t. the open immersion $U\to X$, and the canonical map $$\\{ (x,\varphi),x\in U,\varphi\in \Hom_{local}(\mathcal{O}\_{U, x},R)\\}\to \\{ (x,\varphi),x\in X,\varphi\in \Hom_{local}(\mathcal{O}\_{X, x},R)\\}$$ is an identity map. We have shown $c_U$ is a bijection (defined as $c$ with $U$ replacing $X$). So $c\_U (g\_0)=c\_U(h\_0)$ implies that $g\_0=h\_0$, which in turn implies $g=h$.

The typo above is an example of this issue \ref{https://stackoverflow.com/questions/29014573/mathjax-how-to-deal-with-this-strange-behavior/29040570#29040570}. Note that adding underscore after right curly bracket instantly turned into italc form (you can see the italic button on the tool row becomes highlighted). I added backslash sign before underscore sign to make it looks right in preview. But now in real output it screws up. Let me test again. One backslash: $\mathcal{O}\_{X,x}, \mathcal{O}\_{X,x}$. Correct in preview, probably wrong in real output. No backslash: $\mathcal{O}_{X,x}, \mathcal{O}_{X,x}$. Wrong in preview, probably correct in real output.

@#5879: In stead of giving your own proof, could you comment on what sentence(s) of the current proof should be clarified? I'm a bit surprised you are complaining about the deduction of the general case from the affine case, as that seems to me is explained quite well.

@#5880. Yes. I'm sorry about the mismatch between the editor and what readers get to see. As you say, this has been a problem for a while. We don't quite know how to fix this.

@#5882, denote the functor as $c$, for me the problem of confusion is that the relation between $c$ restricted to open affine $U$ and $c$ on $X$ is not clear. For example, before we prove that $U\hookrightarrow X$ is a monomorphism in the category of schemes in tag 01L7 (so for this tag not proved yet), we cannot use the fact $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ injects into $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$. So in this case I prefer we emphasis the difference between the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ and the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$. Actually I would drow a square for it if I know how to draw it in this editor.

And we don't really need the injection in this proof. For the shortest suggestion on the proof, I would add something like:

We proved every element of $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$ can be lifted to an element in $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ for some open affine $U$. Providing $c$ is functorial. Then the injectivity of $c$ on $X$ reduces to the injectivity of $c$ on $U$, which is already proved.

@#5882: OK, I think you are really saying the following fact is being used silently.

Fact: If $U \subset X$ is an open subspace of a locally ringed space, then the inclusion morphism $U \to X$ is a monomorphism in the category of locally ringed spaces.

I agree it is being used. The Fact follows immediately from Lemma 26.3.5 whose proof is omitted. If more people complain, then I will consider changing the proof. Or maybe somebody could edit the latex file and write a better proof of this lemma (maybe using the aforementioned lemma) and submit it to stacks.project at gmail.com. Thanks!