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I am reading Olsson's book Algebraic Spaces and Stacks, and I am somewhat stuck in understanding one aspect of Example 1.2.4 (in page 10).

The goal is find a morphism of schemes $f: X\to Y$ which is quasi-compact and locally of finite presentation but not quasi-separated. Here is the construction. We let $Y =\operatorname{Spec} k[x_1, x_2, ...]$ denote the infinite affine space, and let $z$ be the closed point of $Y$ corresponding to the origin (where all $x_i=0$). Let $U=Y\setminus \{z\}$ is an open set of $Y$. It can be easily checked that $U$ is not quasi-compact.

Now, let $X$ be the scheme obtained by taking two copies of $Y$, and glueing them along the open set $U$. We can imagine $X$ to be an infinite affine space with "doubled" origin. Let's call $X_1, X_2\subset X$ the two subsets that we glued together (so $X_i\cong Y$). Consider the morphism $f: X\to Y$ which just restricts to the identity on each $X_i$. Then Olsson claims that $X_{1}\times_{Y} X_{2}\cong Y$ (which I kind of understand) and that the following diagram is a Cartesian diagram: $$ \require{AMScd} \begin{CD} U @>>> X_1\times_{Y} X_2\\ @VVV @VVV\\ X @>\Delta>> X\times_{Y} X \end{CD} $$ I understand that this gives the desired conclusion: indeed, if $f: X\to Y$ were quasi-separated, then the diagonal $X\to X\times_{Y} X$ would be quasi-compact, and hence its base-change $U\to X_{1}\times_{Y} X_{2}=Y$ would be quasi-compact, which is false. So my questions are:

1) How are we supposed to visualize the map $\Delta: X\to X\times_{Y} X$? If $X$ is the infinite affine space with doubled origin, what is the corresponding geometric picture for $X\times_{Y} X$? I am hoping this would clarify to me what $\Delta$ is.

2) Why is the above diagram cartesian?

Thanks!

Remark: There is already another post in MSE that shows $X$ is not quasi-separated as a scheme, i.e. the map $X\to \operatorname{Spec}(\mathbb{Z})$ is not quasi-separated. But my questions above still stand.

Prism
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    It looks like the fact that you are working with the infinite affine space is irrelevant for your question(s). If you can answer them for $\mathbb{A}^n$ (or even $\mathbb{A}^2$) then you should be able to understand this case as well. The need to work with infinite affine space is only to have a non-quasicompact map on the top row of your square. – User3773 Sep 02 '17 at 16:12
  • @Cla Yes, you are right. Let's say we are dealing with the case when $Y=\mathbb{A}^{2}$. I am guessing that in the cartesian diagram above, when we pull-back the diagonal morphism $X\to X\times_{Y} X$ along the inclusion (?) $X_{1}\times_{Y} X_{2} \to X\times_{Y} X$, we should get $X_1\cap X_2 \mapsto X$, and since $X_1\cap X_2 = U$, this explains why we have $U\to X$. Is everything I said correct? Feel free to write an answer explaining your point of view when $Y=\mathbb{A}^2$ :) I just want to get a better intuition as to why fiber product above corresponds to intersection $X_1\cap X_2$. – Prism Sep 02 '17 at 17:26
  • Yes, that should be alright. Set-theoretically, it is quite clear (just see explicitly which are the closed points) that $X\times_{X\times_{Y}X}(X_1\times_Y X_2)=U$; if you want a scheme-theoretic identification, then just break the fibre product up in affine pieces (here is where working with $\mathbb{A}^2$ makes the difference!). In this way you get the commutativity of the diagram by definition. – User3773 Sep 02 '17 at 18:01
  • Thanks! I still have one question. Let's do the case when $Y=\mathbb{A}^1$ to make it simpler. In this case, the corresponding $X$ is the usual affine line with doubled origin. Could you tell me what is $X\times_{Y} X$ just as a set? It seems to me that it is an affine line with $4$ origins. – Prism Sep 02 '17 at 18:21
  • Actually I don't know why I stopped at $\mathbb{A}^2$, but yeah $\mathbb{A}^1$ is even better :) At first view $X\times_{\mathbb{A}^1}X$ looks like four copies of $\mathbb{A}^1$ glued "via the identity" and hence yes I'd say it is the line with four origins. – User3773 Sep 02 '17 at 20:54
  • I believe taking the infinite affine space is needed to ensure $U$ is not quasi-compact. Else you will get something non-separated but quasi-separated. For example, in the case $\mathbb{A}^1$, the scheme $U$ is just $\mathbb{G}_m$, which is affine, hence qc. – Maanroof May 19 '22 at 14:03

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