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.

Any comments are welcome.