This is related to [another question](https://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes).


I've found many remarks that the category of schemes is not cocomplete. The category of locally ringed spaces is cocomplete, and in some special cases this turns out to be the colimit of schemes, but in other cases not (which is, of course, no evidence that the colimit does not exist). However, I want to understand in detail a counterexample where the colimit does not exist, but I hardly found one. In FGA explained I've found the reference, that Example 3.4.1 in Hartshorne, Appendix B is a smooth proper scheme over $\mathbb{C}$ with a free $\mathbb{Z}/2$-action, but the quotient does not exist (without proof). To be honest, this is too complicated to me. Are there easy examples? You won't help me just giving the example, because there are lots of them, but the hard part is to prove that the colimit really does not exist.



Colimits of schemes
**Edit:** [BCnrd](https://mathoverflow.net/users/3927/bcnrd) gave a proof in the comments that this example works, so I've edited in that proof.


A ~~possible~~ proven 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, 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.~~


If a coequalizer $P$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $P$. 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.




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


Here are some questions that might be helpful to answer:



> 
> 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?](https://mathoverflow.net/questions/63/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 $Sch\to Set$ *is not* corepresented by a scheme?
> 
> 
> 


