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3 inserted Brian's proof

Edit: BCnrd gave a proof in the comments that this example works, so I've edited in that proof.

## A possibleproven example

I suspectthere There is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, I think that the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.

One thing I can prove is that if

If a coequalizer $X$ P$does exist, then no two closed points of$\mathbb A^1\sqcup \mathbb A^1$map to the same point in$X$. P$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.

On the other hand, let $U$ be an affine open around the image of the generic point in $P$. $U$ has dense open preimages $V$ and $V'$ in both affine lines. Let $W=V\cap V'$ inside the affine line, so we have two maps from $W$ to the affine $U$ which coincide at the generic point of $W$, and hence are equal (as $U$ is affine). In particular, the two maps from affine line to categorical pushout $P$ coincide at each "common pair" of closed points of the two copies of $W$, contradicting the previous paragraph.

Edit: The questions below are no longer relevant, but I'd like to leave them there for some reason.

What are some ways to determine that a functor $Sch^{op}\to Sch\to Set$ is not corepresented by a scheme?

2 added 114 characters in body

## A possible example

I suspect there is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, I think that the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.

I don't have a proof, but I thought other people might have ideas if I posted this here.

One thing I can prove is that if a coequalizer $X$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $X$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.

If the coequalizer above does exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective?

(see the related question Can a coequalizer of schemes fail to be surjective?)

Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of separated schemes equal to $\mathbb A^1$? (probably)

What are some ways to determine that a functor $Sch^{op}\to Set$ is not corepresented by a scheme?

1

## A possible example

I suspect there is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, I think that the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.

I don't have a proof, but I thought other people might have ideas if I posted this here.

One thing I can prove is that if a coequalizer $X$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $X$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.

If the coequalizer above does exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective?
Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of separated schemes equal to $\mathbb A^1$? (probably)