The Family Fibration of Sets: A Fibration with Re-indexed Identities as Cartesian Morphisms

To formally prove that the family fibration of sets is a fibration where the Cartesian morphisms are re-indexed identities, we will proceed in four steps:

1. Define the family fibration of sets.
2. Define the concepts of a fibration and a Cartesian morphism in category theory.
3. Demonstrate that the family fibration satisfies the condition for being a fibration by constructing Cartesian liftings.
4. Characterize these Cartesian morphisms and show they are precisely the "re-indexed identities."

1. The Family Fibration of Sets

The family fibration of sets describes how families of sets are structured over the category of sets.

• Base Category: The base category is Set, the category where objects are sets and morphisms are functions between sets.

• Total Category: The total category, which we will call Fam(Set), is defined as follows:

  • Objects: An object is a pair $(I,A)$, where $I$ is a set (the indexing set) and $A$ is a function $A:I\to \mathrm{Ob}\left(\mathrm{Set}\right)$, which assigns a set $A\left(i\right)$ to each element $i\in I$. This is an $I$-indexed family of sets.
  • Morphisms: A morphism from an object $(J,B)$ to $(I,A)$ is a pair $(f,\phi )$, where $f:J\to I$ is a function, and $\phi$ is an indexed family of functions ${\phi }_j:B\left(j\right)\to A\left(f\right(j\left)\right)$ for every $j\in J$.

• Projection Functor: There is a projection functor $p:\mathrm{Fam}\left(\mathrm{Set}\right)\to \mathrm{Set}$ that forgets the family structure:

  • On objects: $p(I,A)=I$.
  • On morphisms: $p(f,\phi )=f$.

2. Fibrations and Cartesian Morphisms

In category theory, a fibration (or Grothendieck fibration) is a functor $p:E\to B$ that has a special lifting property.

• Fibration: A functor $p:E\to B$ is a fibration if for every object $X$ in $E$ and every morphism $f:J\to p\left(X\right)$ in $B$, there exists a Cartesian lifting of $f$ to a morphism $g:Y\to X$ in $E$ for some object $Y$.
• Cartesian Morphism: A morphism $g:Y\to X$ in $E$ is called Cartesian over a morphism $f=p\left(g\right):p\left(Y\right)\to p\left(X\right)$ in $B$ if it satisfies a specific universal property. For any other morphism $h:Z\to X$ and any morphism $m:p\left(Z\right)\to p\left(Y\right)$ in $B$ such that $f\circ m=p\left(h\right)$, there exists a unique morphism $h\prime :Z\to Y$ in $E$ such that $p(h\prime )=m$ and $g\circ h\prime =h$. This property essentially states that $g$ is a "universal" arrow from the fiber over $p\left(Y\right)$ to the object $X$.

3. Proving the Family Fibration is a Fibration

To prove that $p:\mathrm{Fam}\left(\mathrm{Set}\right)\to \mathrm{Set}$ is a fibration, we must show that for any object $(I,A)$ in Fam(Set) and any function $f:J\to I$ in Set, a Cartesian lifting exists.

Let's construct this lifting. Our goal is to find a Cartesian morphism $g$ in Fam(Set) such that $p\left(g\right)=f$ and its codomain is $(I,A)$.

1. Construct the Domain Object: We define a new object $(J,A\circ f)$ in Fam(Set). This object consists of the indexing set $J$ and a new family of sets $A\circ f:J\to \mathrm{Ob}\left(\mathrm{Set}\right)$ which is the re-indexing of the family $A$ along the function $f$. For each $j\in J$, the set is $(A\circ f)\left(j\right)=A\left(f\right(j\left)\right)$.
2. Construct the Lifting Morphism: We propose the following morphism as our Cartesian lifting:
   $$g=(f,\mathrm{id}):(J,A\circ f)\to (I,A)$$
   Here, $\mathrm{id}$ represents the family of identity functions ${\mathrm{id}}_{A\left(f\right(j\left)\right)}:(A\circ f)\left(j\right)\to A\left(f\right(j\left)\right)$ for each $j\in J$. This morphism is what we term a re-indexed identity. Clearly, $p\left(g\right)=p(f,\mathrm{id})=f$, so it is a valid lifting.
3. Verify the Cartesian Property: Now, we must show that $g=(f,\mathrm{id})$ is Cartesian. We use the universal property defined above:
   • Let $h=(k,\theta ):(K,C)\to (I,A)$ be any other morphism in Fam(Set).
   • Let $m:K\to J$ be a function in Set such that $f\circ m=k$.
   • We need to find a unique morphism $h\prime :(K,C)\to (J,A\circ f)$ such that $p(h\prime )=m$ and $g\circ h\prime =h$.
   Let's define $h\prime =(m,\alpha )$. The condition $p(h\prime )=m$ is satisfied by construction. Now we need to satisfy $g\circ h\prime =h$.
   The composition $g\circ h\prime$ is $(f,\mathrm{id})\circ (m,\alpha )$. By the rules of composition in Fam(Set), this is: $$(f\circ m,\beta )$$ where ${\beta }_x={\mathrm{id}}_{A\left(f\right(m\left(x\right)\left)\right)}\circ {\alpha }_x={\alpha }_x$ for each $x\in K$.
   So, $g\circ h\prime =(f\circ m,\alpha )$. For this to equal $h=(k,\theta )$, we must have:
   • $f\circ m=k$, which is true by our initial assumption.
   • $\alpha =\theta$.
   This uniquely determines $\alpha$. The components of $\alpha$ must be ${\alpha }_x={\theta }_x$. Let's check that the domains and codomains are compatible.
   • For $h\prime =(m,\alpha )$, ${\alpha }_x$ must map $C\left(x\right)\to (A\circ f)\left(m\right(x\left)\right)=A\left(f\right(m\left(x\right)\left)\right)$.
   • For $h=(k,\theta )$, ${\theta }_x$ maps $C\left(x\right)\to A\left(k\right(x\left)\right)$.
   Since $k=f\circ m$, the codomains $A\left(f\right(m\left(x\right)\left)\right)$ and $A\left(k\right(x\left)\right)$ are the same set. Therefore, setting $\alpha =\theta$ is valid.
   We have found a unique morphism, $h\prime =(m,\theta )$, that satisfies the required conditions. Thus, the morphism $g=(f,\mathrm{id})$ is Cartesian. Since we can construct such a Cartesian lifting for any object and any base morphism, $p:\mathrm{Fam}\left(\mathrm{Set}\right)\to \mathrm{Set}$ is a fibration.

4. Characterizing the Cartesian Morphisms

We have shown that "re-indexed identities" of the form $(f,\mathrm{id}):(J,A\circ f)\to (I,A)$ are Cartesian. Now we prove the converse: any Cartesian morphism is of this form, up to isomorphism.

Let $(f,\phi ):(J,B)\to (I,A)$ be an arbitrary Cartesian morphism in Fam(Set) over $f:J\to I$.
We already know that $(f,\mathrm{id}):(J,A\circ f)\to (I,A)$ is also a Cartesian lifting of $f$ with the same codomain $(I,A)$.

A fundamental property of fibrations is that any two Cartesian liftings of the same morphism with the same codomain are uniquely isomorphic in the total category via a vertical isomorphism (an isomorphism that projects to an identity morphism in the base category).

Therefore, there exists a unique vertical isomorphism $v:(J,B)\to (J,A\circ f)$ such that $(f,\mathrm{id})\circ v=(f,\phi )$.

• For $v$ to be vertical, $p\left(v\right)$ must be an identity function, so $p\left(v\right)={\mathrm{id}}_J$. This means $v$ has the form $({\mathrm{id}}_J,\alpha )$.
• For $v$ to be an isomorphism, each component ${\alpha }_j:B\left(j\right)\to (A\circ f)\left(j\right)$ must be an isomorphism (i.e., a bijection, since we are in Set).
• Let's compute the composition: $(f,\mathrm{id})\circ v=(f,\mathrm{id})\circ ({\mathrm{id}}_J,\alpha )=(f\circ {\mathrm{id}}_J,\alpha )=(f,\alpha )$.
• Equating this with $(f,\phi )$ gives $(f,\alpha )=(f,\phi )$, which implies $\alpha =\phi$.

This shows that the family of morphisms $\phi$ must be the same as the family of isomorphisms $\alpha$. Therefore, any Cartesian morphism $(f,\phi ):(J,B)\to (I,A)$ must consist of a family of isomorphisms ${\phi }_j:B\left(j\right)\to A\left(f\right(j\left)\right)$.

In conclusion, the Cartesian morphisms in the family fibration are precisely those morphisms $(f,\phi )$ where $\phi$ is a family of isomorphisms. Such a morphism establishes an isomorphism between the domain object $(J,B)$ and the canonical re-indexed object $(J,A\circ f)$ in the fiber over $J$. Under this identification, the morphism $(f,\phi )$ corresponds to the re-indexed identity $(f,\mathrm{id})$.