Timeline for Unitors and projections in cartesian category
Current License: CC BY-SA 4.0
21 events
when toggle format | what | by | license | comment | |
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Oct 9 at 13:33 | answer | added | anuyts | timeline score: 2 | |
Jul 24 at 16:55 | answer | added | Jean-Baptiste Vienney | timeline score: 2 | |
Dec 19, 2021 at 6:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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Nov 29, 2019 at 23:16 | history | edited | YCor |
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Nov 29, 2019 at 23:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Oct 31, 2019 at 9:25 | comment | added | statusfailed | Right, but when we talk about a cartesian monoidal category, we still have to define a tensor product functor, thereby assigning a particular product object to each $A$ and $B$. When I used $\otimes$ above it was to distinguish that particular object (which is still a categorical product of $A$ and $B$) from other product objects in the category. | |
Oct 30, 2019 at 23:38 | comment | added | Bartosz Milewski | Also, I only mean cartesian product. A general tensor product doesn't necessarily have projections. | |
Oct 30, 2019 at 22:25 | answer | added | statusfailed | timeline score: 1 | |
Oct 30, 2019 at 21:32 | comment | added | statusfailed | Scratch that, I misunderstood! | |
Oct 30, 2019 at 21:11 | comment | added | statusfailed | I'm still having a bit of trouble interpreting this question, because I think it might be ill-posed as written. In particular: If for each object $A$ and $B$ you choose a specific product object $A \otimes B$ - which you must, because $\otimes$ has to be a functor - then you are asking if that object can be a product in more than one way. That is, can the object have multiple sets of valid projections. Is that right? (My problem with your wording is when you say "is it possible to have for the same product a different set of unitors - when you fix the product, you fix $\pi_1$.) | |
Oct 29, 2019 at 22:24 | answer | added | Vlad Patryshev | timeline score: 0 | |
Oct 29, 2019 at 20:07 | comment | added | Bartosz Milewski | Let me add some motivation. In string diagrams we ignore unitors. So, for instance, if we have a morphism $f \colon A \to B$, the same diagram describes $f \circ \lambda_A$ and $\lambda_B \circ (id_1 \times f)$. I can prove this from the universal construction that defines $id_1 \times f$ only if the unitors are equal to the corresponding projections. | |
Oct 29, 2019 at 1:40 | comment | added | Bartosz Milewski | It's not even clear that $\pi'_1$ and $\pi'_2$ define a product. They may not satisfy the universal condition. | |
Oct 28, 2019 at 19:14 | comment | added | statusfailed | A product is both the object and the projections, so once you fix a particular categorical product object as the tensor product, then you also fix the projections - so would it be fair to rephrase this question as follows? Say $(P, \pi_0, \pi_1)$, is a product of $A$ and $B$, is there another product $(P, \pi'_0, \pi'_1)$ with $\pi_0 \neq \pi'_0$ or $\pi_1 \neq \pi'_1$ In other words, once you pick the object, do the projections become unique? | |
Oct 28, 2019 at 18:47 | history | asked | Bartosz Milewski | CC BY-SA 4.0 |