Can global failure of Extensionality in fragments of NFU permit existence of singleton relation set?

Question

Let $SF$ be the schema of stratified comprehension.

Take the theory $SF + Infinity + Choice + \text {Extensionality fails everywhere}$

Is the following consistent with this theory?

$\exists \iota \forall x (x \in \iota \leftrightarrow \exists a (x = \langle a, b \rangle \land \forall m (m \in b \leftrightarrow m=a)))$

In English there exists a global singleton relation set.

It is known that this theory can interpret $NFU + Infinity + Choice$ because SF can interpret NFU a proof due to Marcel Crabbe`. An equivalent proof of NFU being interpretable form SF is present here 15-17.

Answer 1

No.

Let s be the relation whose elements are all the pairs (x,y) such that x is the only element of y [the set the poster postulates].

Now let R be the set of all x such that for all y such that (x,y) E s, y is not a subset of x. This definition is stratified: if s exists, R must exist. This set R is the Russell class, so the poster's postulated relation (it isn't a function as I originally thought, because of the failure of extensionality) cannot exist.

This is somewhat obscured in the poster's formulation, as without extensionality it is unclear what is meant by (a,b): this can be dispelled by defining x = (a,b) as meaning "x has two elements, one of which has a as its only element and one of which has a and b as its only elements". A function is then defined as a set of ordered pairs in this sense in which any two pairs with the same first element also have the same second element, factoring out the fact that there may be many implementations of any particular pair.