Quotient sets in HoTT

Question

Let $A$ be a set in HoTT. A relation $R$ on $A$ is an embedding $r: R \to A\times A$. The quotient set $A/R$ of $A$ by $R$ is the higher inductive type with the following constructors:

1. $c: A \to A/R$
2. $h: \prod_{x: R} c(r_1(x)) = c(r_2(x))$
3. $p: \mathrm{isSet}(A/R)$

My question is about condition $3$, can we drop it and prove that $A/R$ is a set?

In an $\infty$-topos $C$ with $1$-site, for any representation of groupoid $R\to U\times U$ the quotient stack $[U/R]$ is a sheaf of setoids if and only if $R\to U\times U$ is an injection. So I think we should be able to prove this.

Edit: I find out where I went wrong. In the definition of quotient stack it is in fact homotopy $2$-colimit. In other words, it contains higher constructors.

Answer 1

We cannot drop condition (3), at least in the presence of a univalent universe $\mathcal{U}$.

First, a type theoretic proof: Let $A = \{a,b\}$ and $R = \{(a,b), (b,a)\}$. Then we can define a map from the quotient type $A/R$ into $\mathcal{U}$ by giving maps $c' : A \to \mathcal{U}$ and $h' : \prod_{x : R} c'(r_1(x)) = c'(r_2(x))$, which we take to be

1. $c' : a,b \mapsto \mathbb{B}$, the booleans $\mathbb{B} = \{T,F\}$
2. $h' : (a,b) \mapsto \text{refl}_\mathbb{B}$ and $(b,a) \mapsto \text{ua}(\text{not})$

Now if $A/R$ were a set we would know that $h(a,b) = h(b,a)$ and so by applying the function $A/R \to \mathcal{U}$ we just built we would learn $\text{refl}_\mathbb{B} = \text{ua}(\text{not})$ and now transporting $T$ along this path gives $T = F$.

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Next, a less formal argument: The point is that our constructor $h$ can identify two points in multiple ways. In the above example, we added two paths from $a$ to $b$ in $A$ so that in the free type with these paths (read: in $A/R$) the two paths have to be different. That means we have a circle between $a$ and $b$, which is not a set.

I'm not sure what you mean by your final comment, since if $U$ is any groupoid then $R = \emptyset$ injects into $U \times U$ and the quotient $[U/R]$ is just $U$ again, which has no reason to be a setoid. Moreover, we probably want to be working with sheaves of sets rather than sheaves of (1-)gropuoids, since at the start of the problem $A$ was supposed to be an (internal) set.

So if you wanted to make this kind of semantic argument, you would want to find conditions for which the quotient stack of two sheaves of sets $R \to U \times U$ is again a sheaf of sets. Intuitively this means you want to add a unique path between any two points, and there's many ways to ensure this. The most obvious approach is to just assert that paths are unique by adding condition (3) back in. If you want to do it by hand, you might start by assuming that $R$ is an equivalence relation and adding extra constructors asserting that $h(x,y) = h(y,x)^{-1}$ and $h(x,y) \cdot h(y,z) = h(x,z)$. It's possible that this works, but it's also possible that you would need to add constructors for "even higher homotopies" between these. I haven't thought very hard about it, unfortunately.

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I hope this helps ^_^