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.