Universal Family of a Fine Moduli Space

I'm reading up on fine moduli spaces and I'm having difficulty seeing how every family over a scheme $B$ is the pullback of the universal family along a unique morphism. In fact, I'm not sure what this means.

To make my question more precise, I'll use the notation of Harris and Morrison, If $F$ is a moduli functor representable by a scheme $M$, let $\Psi: Mor(-,M) \to F$ be the corresponding natural isomorphism. Pulling back the identity on $M$, $1_M$, we get a family in $F(M)$, $\mathbf{1}: U\to M$. Let $\phi: D\to B$ be a family in $F(B)$. How does one realize this family as the pullback of $U$ via $\Psi(\phi)$?

Furthermore, I've seen the claim that $D\cong B\times_M U$. Harris and Morrison claim that there is a fibre product diagram

$\begin{array}{ccc} D &\rightarrow& U \\ \phi \downarrow && \downarrow \mathbf{1} \\ B&\xrightarrow{\Psi(\phi)}& M\end{array}$

but what is the top morphism and why is this a fibre product?

• I think the "moduli functor" bit adds an extra layer of confusion at this stage. Do you understand the Yoneda lemma? In particular you want to understand how to reconstruct an isomorphism $h_X \simeq F$ of contravariant functors $\mathfrak C \to \mathfrak{Set}$ from the data of just an element $u \in F(X)$. – Hoot Aug 23 '16 at 0:19
• Yes, define $\alpha: h_X\cong F$ by $\alpha_X: 1_X\mapsto u$. By naturality, this determines the entire isomorphism. But how does that help me? – Rdrr Aug 23 '16 at 0:26
• Okay, I dug up my copy of the book. One problem you're having is that they never really lay out what's expected of $F$. I think it should be demanded that a map of "base schemes" $B \to B'$ should induce a map of families via pullback. I doubt it follows from representability and in any case the most important functor in the book isn't representable! – Hoot Aug 23 '16 at 0:55
• Say, we did have the change of base scheme condition you said. How does a given family become the pullback of the universal family? – Rdrr Aug 23 '16 at 17:22
• Well Yoneda says that the class of $[\phi]$ is the same as $F(\mathbf{1}) = [\mathscr{C} \times_{\mathscr M} B \to B]$. So they're isomorphic over $B$. Maybe I'm not understanding you. – Hoot Aug 24 '16 at 1:55

As pointed out in the comments, a functor acts not only on objects but also on morphisms. So in order to define your functor $$F$$, you have to specify not only a set $$F(B)$$ for each scheme $$B$$, but also a map $$F(\varphi): F(B') \to F(B)$$ for each morphism $$\varphi: B \to B'$$ of schemes. (This is assuming $$F$$ is contravariant.)

When $$F$$ is a moduli functor so that $$F(B)$$ classifies objects over $$B$$ up to some equivalence, this leads naturally to the fiber product. In order to specify $$F(\varphi)$$, for each class of objects $$[X]$$ over $$B'$$, we must specify some class of objects over $$B$$. In other words, we have the diagram

$$\hspace{6cm}$$

that we need to complete. The instinctive choice is the fiber product $$X \times_{B'} B$$, and so we define $$F(\varphi)([X]) = [X \times_{B'} B]$$.

$$\hspace{4cm}$$

As pointed out in the comments, the proof that every class of objects can be realized as the pullback of the universal object is the diagram chase in the proof of Yoneda's lemma. Let $$h_M = \operatorname{Mor}(\, \cdot \,, M)$$ and, using your notation, let $$\Psi: h_M \overset{\sim}{\longrightarrow} F$$ be the natural isomorphism. Given a scheme $$B$$ and an object $$X \in F(B)$$, then there exists a unique $$\varphi \in h_M(B) = \operatorname{Mor}(B,M)$$ such that $$\Psi_B(X) = \varphi$$. Let $$\DeclareMathOperator{\id}{id} U = \Psi_M(\id_M)$$ be the universal object over $$M$$. The very trivial fact that $$\id_M \circ \varphi = \varphi$$ then shows via the following diagram chase that $$X$$ is the pullback of $$U$$ along $$\varphi$$.

I haven't read Harris and Morrison's book carefully, but it looks like they breeze through the background of moduli problems. A more detailed resource that provides a careful treatment of moduli problems is Dinamo Djounvouna's masters thesis, The Construction of Moduli Spaces and Geometric Invariant Theory.

• Please use MathJax. For some basic information about constructing commutative diagrams at this site see e.g. MathJax tutorial and quick reference. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – GNUSupporter 8964民主女神 地下教會 Dec 13 '18 at 10:03
• @GNUSupporter8964民主女神地下教會 You can’t be serious. Have you ever tried making a commutative diagram with AMScd? – André 3000 Dec 13 '18 at 10:52
• I've made use of AMScd to answer questions. The demo in the tutorial is quite simple. – GNUSupporter 8964民主女神 地下教會 Dec 13 '18 at 11:15
• @GNUSupporter8964民主女神地下教會 AMScd is very primitive. It's not possible to make arrows with different types of heads or tails, such as the \mapsto arrow used in my third diagram. There isn't even support for diagonal arrows. There are many meta posts about AMScd's shortcomings, such as 1, 2, 3. – André 3000 Dec 13 '18 at 16:50
• @GNUSupporter8964民主女神地下教會 Let me get this straight. Each time I make a diagram, you expect me to know in advance whether I will need any dashed, curved, or diagonal arrows or any special arrowheads or -tails—in short, any of the many features not supported by AMScd. Because if I start making a diagram using AMScd and then suddenly realize I need one of these unsupported features, I’ll have to open up my TeX editor and basically start over from scratch. No, that’s ridiculous. I’m going to use a package that has the tools I need, not some primitive relic that can’t even draw a diagonal arrow. – André 3000 Dec 13 '18 at 21:19