0
$\begingroup$

In book: https://math.mit.edu/~dspivak/teaching/sp18/7Sketches.pdf 2.87(b) & answer of 2.104(1) assume that left adjoint preserve join applicable on empty set, i.e. $ \left( v \otimes \bigvee_\left(a \in A \right)a\right) \cong \bigvee_\left(a \in A\right)(v \otimes a)$ where $A \subseteq V$.

$\endgroup$
1
$\begingroup$

I think it's not applicable on $\emptyset$, as counter example: $ \left(\left[1, \infty\right], \le, \times, \div\right) $ is symmetric monoidal preorder that is closed.
But $ 2 \times \bigvee_\left(a \in \emptyset\right) a = 2 \neq 1 = \bigvee_\left( a \in \emptyset\right) (2 \times a )$

Please advise if I have any misunderstanding here?

$\endgroup$
2
  • 1
    $\begingroup$ I don't see that this monoidal structure is closed, because your "Hom" operation, division, can produce numbers smaller than $1$. $\endgroup$ – Andreas Blass Apr 14 at 3:27
  • $\begingroup$ Thank you @AndreasBlass. I see it's not closed $\endgroup$ – Shark Xu Apr 14 at 5:52
1
$\begingroup$

Yes, the initial object is the empty colimit and is thus preserved by all left adjoints. In the context of a poset this says that all left adjoints preserve the bottom element.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.