Timeline for Why does tensor product in Ab(V) require colimits in V?
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11 events
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| Dec 5, 2012 at 3:40 | comment | added | ziggurism | I guess I already worked this out for myself, with help from Zhen, but chapter 3 of Steve Awodey's category theory online lecture notes have a description of how to construct a presentation out of a coequalizer of the free group on relations. | |
| Nov 29, 2012 at 23:37 | comment | added | ziggurism | Then $A\otimes B=\text{Coeq}(\lambda,\rho)$. So if all the limits and colimits used in this construction exist, the tensor product exists. | |
| Nov 29, 2012 at 23:35 | comment | added | ziggurism | Now to restate it without "global elements". Set $\lambda=UF(\mu_A\times\text{id}_B\coprod \text{id}_A\times\mu_B),$ where $\mu_A\colon A\times A\to A$ is the multiplication morphism. And $\rho=UF((\text{pr}_{\hat{2}}+\text{pr}_{\hat{1}})\coprod(\text{pr}_{\hat{3}}+\text{pr}_{\hat{2}}))$, where $\text{pr}_{\hat{2}}=\text{pr}_1\times\text{pr}_3\colon X\times Y\times Z\to X\times Y$ is the projection which drops the second component. I've assumed here that $\textbf{Ab}(\mathcal{V})$ is preadditive (so I can take the sums $\text{pr}_{\hat{2}}+\text{pr}_{\hat{1}}$ etc.) (cont) | |
| Nov 29, 2012 at 23:25 | comment | added | ziggurism | @Zhen: I guess it must be obvious how to make this explicit if you have experience with generators and relations, but I found this situation confusing, maybe because of the large number of generators. After a while of staring at it, I think I got it. It should model this situation: $R=A\times A\times B\coprod A\times B\times B.$ Then we have $\lambda,\rho\colon UF(R)\rightrightarrows UF(A\times B)$ with $\lambda(a_1,a_2,b_1)=(a_1+a_2,b_1)$,$\lambda(a_3,b_2,b_3)=(a_3,b_2+b_3)$,$\rho(a_1,a_2,b_1)=(a_1,b_1)+(a_2,b_1)$, and $\rho(a_3,b_2,b_3)=(a_3,b_2)+(a_3,b_3).$ (cont) | |
| Nov 28, 2012 at 19:55 | comment | added | Zhen Lin | @ziggurism Something like that, but not quite. A morphism $1 \to A$ is called a "global element" (by analogy with what happens in the topos of sheaves) but we need to make sure the tensor product does the right thing to all "elements". A more precise description would say something like, we have a diagram $I \rightrightarrows U F (A \times B)$ in $\mathcal{V}$ tabulating all the equations we want to be true in $A \otimes B$, and then transposing across the adjunction $F \dashv U$, we build $A \otimes B$ as the coequaliser of $F I \rightrightarrows F (A \times B)$ in $\textbf{Ab}(\mathcal{V})$. | |
| Nov 28, 2012 at 18:26 | comment | added | ziggurism | When you say "once we know that $\textbf{Ab}(\mathcal{V})$ has coequalisers ... we can construct tensor products in the usual way", I'm imagining you mean something like: set $A\otimes B=\text{Coker}\{I\to F(A\times B)\}$, where $F$ is the free abelian group object functor (which exists because $\textbf{Ab}(\mathcal{V})\to \mathcal{V}$ is monadic?) and I is some subobject of $F(A\times B)$ generated by elements like $(a+a',b)−(a,b)−(a',b)$, and here "element" means "morphism $1\to A$". And we have 1 and $\times$ since $\mathcal{V}$ is Cartesian. Is that right? | |
| Nov 28, 2012 at 18:00 | vote | accept | ziggurism | ||
| Nov 28, 2012 at 17:59 | comment | added | ziggurism | The explanation for the Hopf algebra axioms that the comonoid and antipode structures are precisely the thing you need to make group objects work is lovely and surprising. Thank you! As for the construction of the tensor product out of reflexive coequalisers, that pretty much went over my head. But the answer to my question is clear enough: tensor objects need not exist, but some extra axioms, including the existence of the right kinds of colimits, will guarantee them. Thank you also for the link to your notes on algebraic theories. They look very useful to me. | |
| Nov 27, 2012 at 23:15 | comment | added | Zhen Lin | Indeed – that's theorem 3.4.11. in my notes. Actually, if we only want representability of the bihomomorphism functor then we don't even need commutativity – see theorem 3.3.17. As far as I can tell we need some condition on the preservation of (regular) epimorphisms in order to prove the theorem. | |
| Nov 27, 2012 at 23:07 | comment | added | Chris Heunen | Re point 2. This in fact works for any commutative monad whose algebra category has coequalisers of reflexive pairs. This is due to Anders Kock in the 1970s, see dx.doi.org/10.1007/BF01304852 and dx.doi.org/10.1007/BF01220868. | |
| Nov 27, 2012 at 22:04 | history | answered | Zhen Lin | CC BY-SA 3.0 |