# Hom-space between Picard stacks

This is from Deligne, La formule de dualité globale, SGA 4, tome 3, Expose XVIII, and I am confused about how the hom-space between Picard stacks is again a Picard stack.

A quick rewind. For a site $$S$$, a stack $$P/S$$ is called a Picard stack, if there is an addition $$+: P\times_S P \to P$$ so that each fibre $$P(U)$$ for $$U \in S$$ is a Picard groupoid. It is then claimed that $$Hom(P_1, P_2)$$ is a Picard stack, but only each $$Hom(P_1, P_2)(U)$$ is given: objects are additive morphisms between $$P_1(U)$$ and $$P_2(U)$$, and morphisms are morphisms between additive functors.

I wonder why the the existence of cartesian lifts are left blank in Deligne's work. Is there a general principle that allows one to fill in this gap?

I tried to at least figure out morphisms between $$F \in Hom(P_1, P_2)(U)$$ and $$G \in Hom(P_1, P_2)(V)$$. My guess is the following. A morphism $$\alpha:F \to G$$ that projects to $$f:U \to V$$ is given by the following data. For each morphism $$\phi: y \to x$$ over $$f$$ in $$P_1$$, associate $$\alpha(\phi):Fy \to Gx$$. The collection of maps $$\alpha$$ satisfy:

1. If $$\psi: y' \to y$$ is an isomorphism in $$P_1(U)$$, $$\alpha(\phi\circ\psi) = \alpha(\phi)\circ F(\psi)$$.
2. If $$\psi: x \to x'$$ is an isomorphism in $$P_1(V)$$, $$\alpha(\psi\circ\phi) = G(\psi)\circ \alpha(\phi)$$.
3. $$\alpha$$ completes a square between $$F(y_1 + y_2) \cong Fy_1 + Fy_2$$ and $$G(x_1 + x_2) \cong Gx_1 + Gx_2$$.

If this guess actually works, my question would be why there exists at least one such $$\alpha$$ that projects to $$f$$, and why any such $$\alpha$$ is Cartesian?

## 1 Answer

Be careful that $$Hom(P_1,P_2)(U)$$ is not additive functors $$P_1(U)\to P_2(U)$$, but Picard stack morphisms $$P_{1\mid U}\to P_{2\mid U}$$, it's exactly the same formula as for hom-sheaves.

In particular, if you take the "stacks as sheaves of groupoids" perspective rather than the fibered point of view, it's now clear how to restrict such a thing to $$V$$ when you have a morphism $$V\to U$$, because you can simply further restrict to $$V$$ something that was a morphism between restrictions to $$U$$.

• Oh, thanks. I got it now! – Frid Fu Jan 21 at 22:12