# Diagonal is representable then composition is representable

Let $$\mathcal{X}$$ be a stack over $$S$$ i.e., a stack over category of schemes over $$S$$ (which we denote by $$Sch/S$$) which comes with a functor $$\mathcal{X}\rightarrow Sch/S$$. Consider the diagonal map of stacks $$\mathcal{X}\rightarrow \mathcal{X}\times_{S}\mathcal{X}$$. It is known that, if this diagonal map is representable, then, for any object $$X$$ of $$Sch/S$$ i.e., an $$S$$-scheme, any map of stacks $$X\rightarrow \mathcal{X}$$ is representable. The category of $$Sch/S$$ is not special here.

Let $$\mathcal{X}$$ be a stack over $$\mathcal{Y}$$ (which comes with a functor $$\mathcal{X}\rightarrow \mathcal{Y}$$). Suppose that the diagonal map $$\mathcal{X}\rightarrow \mathcal{X}\times_{\mathcal{Y}}\mathcal{X}$$ is representable. Then, for any object $$Y$$ of $$\mathcal{Y}$$, any map of stacks $$Y\rightarrow \mathcal{X}$$ is representable.

Let us fix $$\mathcal{Y}$$ to be the category of Manifolds, denoted by $$\text{Man}$$. Let us go one step higher (or some type of relative notation). Suppose we have two stacks over $$\text{Man}$$. Let $$\mathcal{D}\rightarrow \text{Man}$$ and $$\mathcal{C}\rightarrow \text{Man}$$ be two stacks. We are also given a map of stacks $$F:\mathcal{D}\rightarrow \mathcal{C}$$.

I want to conclude about relative notion of representability of maps from an object of $$\text{Man}$$. Something like,

Let $$M$$ be an object of $$\text{Man}$$. If a map of stacks $$p:M\rightarrow \mathcal{D}$$ is representable, then the composition $$F\circ p : M\rightarrow \mathcal{C}$$ is representable.

To conclude this,

• asking for diagonal map $$\mathcal{D}\rightarrow \mathcal{D}\times_{\text{Man}}\mathcal{D}$$ is useless as the map $$M\rightarrow \mathcal{D}$$ is already representable. So, this extra condition would give mostly nothing new.
• asking for diagonal map $$\mathcal{C}\rightarrow \mathcal{C}\times_{\text{Man}}\mathcal{C}$$ is least interesting case. As every map of stacks of the form $$N\rightarrow \mathcal{C}$$ is then representable, so would be the composition $$M\rightarrow \mathcal{D}\rightarrow \mathcal{C}$$. This would have nothing to do with representability of $$M\rightarrow \mathcal{D}$$.

As I am looking for a relative notion of representability, I thought I should impose the condition that the relative diagonal map i.e., the diagonal map $$\mathcal{D}\rightarrow \mathcal{D}\times_{\mathcal{C}}\mathcal{D}$$ (induced from $$F$$) is representable. So, the result I want to prove is

Suppose $$F:\mathcal{D}\rightarrow \mathcal{C}$$ be a map of stacks (over the category $$\text{Man}$$) such that the diagonal $$\mathcal{D}\rightarrow \mathcal{D}\times_{\mathcal{C}}\mathcal{D}$$ is representable. Let $$M$$ be an object of $$\text{Man}$$. If a map of stacks $$p:M\rightarrow \mathcal{D}$$ is representable, then the composition $$F\circ p : M\rightarrow \mathcal{C}$$ is representable.

We are given $$p:M\rightarrow \mathcal{D}$$ is representable and the diagonal $$\mathcal{D}\rightarrow \mathcal{D}\times_{\mathcal{C}}\mathcal{D}$$ is representable. Suppose we are given a map $$q:N\rightarrow \mathcal{C}$$. We want to prove that $$X\times_\mathcal{C}N$$ is a manifold (this is what we mean when we say $$F\circ p:M\rightarrow \mathcal{C}$$ is representable).

As $$M\rightarrow \mathcal{D}$$ is representable, the stack $$M\times_\mathcal{D}M$$ is a manifold. There is an obvious map $$M\times_{\mathcal{D}}M\rightarrow \mathcal{D}\times_\mathcal{C}\mathcal{D}$$ given by $$(m_1,m_2,\alpha:p(m_1)\rightarrow p(m_2))\mapsto (p(m_1),p(m_2),F(p(m_1))\rightarrow F(p(m_2))).$$

As $$\mathcal{D}\rightarrow \mathcal{D}\times_{\mathcal{C}}\mathcal{D}$$ is representable, considering the map of stacks $$M\times_{\mathcal{D}}M\rightarrow \mathcal{D}\times_\mathcal{C}\mathcal{D}$$ (note that $$M\times_{\mathcal{D}}M$$ is a manifold), $$\mathcal{D}\times_{\mathcal{D}\times_{\mathcal{C}}\mathcal{D}}(M\times_{\mathcal{D}}M)$$ is a manifold.

We have $$\mathcal{D}\times_{\mathcal{D}\times_{\mathcal{C}}\mathcal{D}}(M\times_{\mathcal{D}}M) \cong (M\times_{\mathcal{D}}M) \times_{M\times_{\mathcal{C}}M}(M\times_{\mathcal{D}}M)$$ Above result follows from an isomorphism in proposition (page $$11$$) here.

This does not anyways say that the composition $$M\rightarrow \mathcal{D}\rightarrow \mathcal{C}$$ is an atlas. For that, we need to prove $$M\times_{\mathcal{C}}N$$ is a manifold. All we know is that $$M\times_{\mathcal{C}}M$$ is a manifold.