I think that wikipedia article on Algebraic spaces contains a serious content error in the part on the definition of Algebraic spaces as quotients of schemes and I would like to discuss if it is indeed an error or I just missing something. The definition says:

An algebraic space $X$ comprises a scheme $U$ and a

closed(!) subscheme $R \subset U \times U$ satisfying the following two conditions:

- $R$ is an equivalence relation as a subset $U \times U$;
- the two projections $p_i: R \to U$ onto each factor are étale maps

Some authors, such as Knutson, add an extra condition that an algebraic space has to be quasi-separated, meaning that the diagonal map is quasi-compact. One can always assume that $R$ and $U$ are affine schemes. Doing so means that the theory of algebraic spaces is not dependent on the full theory of schemes, and can indeed be used as a (more general) replacement of that theory.

I think that in this definition is a subtle problem hidden, since it is important to distinguish strictly if $U$ *may be always assumed to be affine or not* when we work with the condition that the $R \subset U \times U$ is assumed to be a *closed* subscheme as I will explain below why. So I'm not sure why it can "one can always assume that $R$ and $U$ are affine". That's doesn't matter only in the case if $U$ is moreover separated as we will see in following example:

It is well know that an arbitrary scheme $S$ is canonically an algebraic space. If we want to verify it using the characterization above we have to recognize it as a quotient of schemes, so we need a pair $(U,R)$ representing $S$ and satisfying axioms $1$ and $2$.

For $U$ we set $U:=S$ and for $R$ we need a *closed* subscheme of $U \times U$ and as canonical choice for $R$ might be choosen the *closure* $R:=\overline{\Delta(U)}$ of the image with respect the diagonal map $\Delta: U \to U \times U$. Here I'm not sure if that is a correct choice for $R$ in the case when $S$ is not separated, but it seems to be the only "canonical" choice. And if this choice for $R$ is correct, we have a problem: Why is the restriction of the projection map $p \vert _{\overline{\Delta(U)}}: \overline{\Delta(U)} \to U$ etale? (take the closure bar into account)

Clearly $p \vert _{\Delta(U)}: \Delta(U) \to U$ is etale since $U \cong \Delta(U)$ and $p \circ \Delta= id_U$, but in general if $f: X \to Y$ is a morphism of schemes, $Z \subset X$ an locally closed immersion (but not cl imm) and the restriction $f \vert _Z:Z \to Y$ is etale, then in general $f \vert _{\overline{Z}}: \overline{Z} \to Y$ is not more etale.

Thus there is no reason if we have a scheme $S (=U)$ with $\overline{\Delta(U)} \not \cong \Delta(U)$ for restriction $p \vert _{\overline{\Delta(U)}}: \overline{\Delta(U)} \to U$ to be etale. On the other hand affine schemes are separated and therefore $\overline{\Delta(U)} = \Delta(U)$, but for general scheme $S$ the requirement that $R \subset U \times U$ is closed is stronger than to say one may always assume that $R$ and $U$ are affine schemes. Or do I confuse something here? Is there maybe a better choice for $R$ involved?

In summary the concern of my question can be reduced to two question:

A: Is for arbitrary scheme $S$ the restriction of projection $p \vert _{\overline{\Delta(S)}}: \overline{\Delta(S)} \to U$ etale?

B: If A is true, then B is clear. If A is wrong, why it is sufficient to work with this definition of algebraic spaces with affine schemes in viewpoint the described problemwith closedness of $R$ in $U \times U$?