Jonathan wrote approximately:

> The latter condition actually shows that \\( \cdot \wedge b \\) and \\(b \to \cdot\\) are adjoint, correct?

Right.

> (Of course, \\(\wedge\\) is commutative.)

Right, so \\(\cdot \wedge b = b \wedge \cdot \\). (Mathematicians use \\(\cdot\\) for a slot that's left open to put something in when describing functions.)

> We didn't actually use the \\((a \wedge)\\) and \\((a \to)\\) introduced prior.

I introduced \\(a\\) in a phrase containing "for each element \\(a \in A\\)", so it applies even if that element happens to be called \\(b\\). So, everything is fine: I merely failed to use the variable names in a way that would make the explanation easy to follow. I'll go back and smooth it down.

> The latter condition actually shows that \\( \cdot \wedge b \\) and \\(b \to \cdot\\) are adjoint, correct?

Right.

> (Of course, \\(\wedge\\) is commutative.)

Right, so \\(\cdot \wedge b = b \wedge \cdot \\). (Mathematicians use \\(\cdot\\) for a slot that's left open to put something in when describing functions.)

> We didn't actually use the \\((a \wedge)\\) and \\((a \to)\\) introduced prior.

I introduced \\(a\\) in a phrase containing "for each element \\(a \in A\\)", so it applies even if that element happens to be called \\(b\\). So, everything is fine: I merely failed to use the variable names in a way that would make the explanation easy to follow. I'll go back and smooth it down.