Suppose that f : Y —> X is a morphism of schemes with f locally of finite type and Y affine. Then there exists an immersion Y —> A^n_X of Y into affine n-space over X. See the slightly more general Lemma Tag 04II.

Now suppose that f : Y —> X is a morphism of algebraic spaces with f locally of finite type and Y an affine scheme. Then it is not true in general that we can find an immersion of Y into affine n-space over X.

A first (nasty) counter example is Y = Spec(k) and X = [A^1_k/Z] where k is a field of caracteristic zero and Z acts on A^1_k by translation (n, t) —> t + n. Namely, for any morphism Y —> A^n_X over X we can pullback to the covering A^1_k of X and we get an infinite disjoint union of A^1_k’s mapping into A^{n + 1}_k which is not an immersion.

A second counter example is Y = A^1_k —> X = A^1_k/R with R = {(t, t)} \coprod {(t, -t), t not 0}. Namely, in this case the morphism Y —> A^n_X would be given by some regular functions f_1, …, f_n on Y and hence the fibre product of Y with the covering A^{n + 1}_k —> A^n_X would be the scheme

{(f_1(t), …, f_n(t), t)} \coprod {(f_1(t), …, f_n(t), -t), t not 0}

with obvious morphism to A^{n + 1} which is not an immersion. Note that this gives a counter example with X quasi-separated.

I think the statement does hold if X is locally separated, but I haven’t written out the details. Maybe it is somehow equivalent to X being locally separated?

Perhaps the correct weakening of the lemma that holds in general is that given Y —> X with Y affine and f locally of finite type, there exists a morphism Y —> A^n_X which is “etale locally on X and then Zariski locally on Y” an immersion? (This does not seem to be a very useful statement however… although you never know.)

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