Epimorphisms and 2-isomorphic maps to an algebraic stack

Question

$\DeclareMathOperator\Spec{Spec}$Let $L/K$ be a field extension, and let $\mathcal{M}$ be some moduli stack (for example, the stack of genus $g$ curves).

Let $X$, $X'$ be two objects of $\mathcal{M}$ over $K$, giving us two morphisms $X,X':\Spec K\rightarrow\mathcal{M}$. Suppose their pullbacks $X_L$, $X'_L$ are isomorphic, which is to say that the two composed morphisms $$\Spec L\rightarrow\Spec K\rightrightarrows \mathcal{M}$$ are 2-isomorphic. Now, I sort of want to say that because $\Spec L\rightarrow\Spec K$ is an epimorphism, that $X$, $X'$ must have determined 2-isomorphic morphisms, and hence were already isomorphic (over $K$) in the first place. …But this is obviously wrong (for example, take $\mathcal{M}$ to be the moduli stack of elliptic curves, and $X$, $X'$ to be two nonisomorphic (over $K$) elliptic curves with the same $j$-invariant).

Where exactly is the problem?

For example,

1. Is $p : \Spec L\rightarrow\Spec K$ not an epimorphism in the category of algebraic stacks?
2. Perhaps the right question is — Is $p$ a 2-epimorphism in the 2-category of algebraic stacks? What is a down-to-earth definition of a 2-epimorphism anyway? (nlab was not especially helpful in this regard.)
3. Does the problem that arises in this situation disappear if we assume that $L$, $K$ are both algebraically closed?

Answer 1

Let more generally $f : T \to S$ be a fpqc covering and $X,X' \in \mathcal{M}(S)$ with $f^* X \cong f^* X'$ in $\mathcal{M}(T)$. By definition(*) of a prestack in the fpqc-topology, a necessary and sufficient condition for $X \cong X'$ (inducing the isomorphism $f^* X \cong f^* X'$) is that the following diagram commutes, where $p_1,p_2 : T \times_S T \rightrightarrows T$ are the two projections:

$$\begin{array}{ccc} p_1^* f^* X & \rightarrow & p_1^* f^* X' \\ \downarrow && \downarrow \\ p_2^* f^* X & \rightarrow & p_2^* f^* X' \end{array}$$

Vertically, we have the isomorphisms induced by $f p_1 = f p_2$. Horizontally, we have the isomorphisms induced by $f^* X \cong f^* X'$.

(*) A prestack (resp. stack) $\mathcal{M}$ is defined to be a fibered category such that for every covering $f:T \to S$ the functor $\mathcal{M}(S) \to \mathcal{M}_{\mathrm{descent}}(f)$ into the category of descent data is fully faithful (resp. an equivalence of categories).

Answer 2

$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Aut{Aut}$I think the issue here is that points of algebraic stacks have nontrivial automorphisms. In your example, the issue is that the 2 isomorphism is defined over $L$ but not over $K$, i.e., the automorphism group of that point change its structure upon a field extension. Certainly, for schemes, the map $\Spec L \to \Spec K$ is an epimorphism, since the automorphism group is trivial so it's defined over the integer. Now, if I want to remove that issue, I may want to assume that the automorphism group bears a trivial action of $\Aut_K(L)$. For nice algebraic stacks, the stabilizer groups are algebraic groups, so if $K$ is algebraically closed, then its structure is already stable under any field extension, i.e., its structure will not change anymore upon field extensions, this reduce the problem to the case where $L/K$ is algebraic.

Also, note that in the general case, where the source is no longer the spectrum of a field but only an algebraic stacks, then we can test whether two morphisms are the same on smooth presentations of the source algebraic stacks, and this reduces the problem to the case where the source is an affine scheme. Then we need to study the Galois action on both the set of 1-morphisms and also 2-morphisms. I guess that is the rough idea.