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following public demand, I added to tensor product of chain complexes a detailed elementary discussion of the tensor product $I_\bullet \otimes I_\bullet$ of the (normalized) chain interval with itself, and how it gives chains on the cellular square: in Square as tensor product of interval with itself.
If you give me a pointer to the code on MO, I’ll try to convert it.
The MathOverflow code is this one:
$$P(i,j;k) = \begin{cases} \mathbb{Z} & \mathrm{if\,\,} i+j=k, \\
\mathbb{Z} \oplus \mathbb{Z} & \mathrm{if\,\,} i+j=k+1, \\
\mathbb{Z} & \mathrm{if\,\,} i+j=k+2, \\
0 & \mathrm{else}.
\end{cases}$$
Hi Dmitri, I am not sure what you were attempting, but I have modified the entry now to use more or less exactly this code, and it works fine. Double dollars should have a blank line before and after; maybe that was the cause of the problems you ran into. Itex2MML didn’t seem to like the \,
inside the \mathrm
, so I just added some space in a different way.
Adds a section about the canonical forgetful functor $Ch(I,-)$ corresponding to the closed elements of degree 0, not all elements of degree 0.
I was somewhat confused about this and the internal hom as documented here. It was suggested I may edit internal hom of chain complexes to save future people from tripping up the same way I did. I think this part fits better here and I will make another edit about the implications to internal hom of chain complexes there next.
I’m not sure the paragraph describing the canonical forgetful functor really belongs here. I didn’t find a page or mention of it.
But why are we doing this in tensor production of chain complexes now? I’ll copy this to category of chain complexes.
Ok. I thought tensor product of chain complexes was the natural place, but maybe that page is more about the construction, with further discussion and properties of the resulting closed structure belonging in category of chain complexes?
It’s maybe not so important. Whatever works for you.
Also, is there a reason this is only defined for $\mathcal{A} = R\text{Mod}$? If $\mathcal{A}$ is a closed $Ab$-category with countable coproducts and products and we replace the direct sum in the construction with the coproduct, does something go wrong? Specifically $\otimes_{\mathcal{A}}$ will commute with the coproduct, which I think is not clear for the notions at direct sum. Is there just a reference/proof missing or do we need more exactness properties for everything to be well-behaved?
Oh, I had only replaced “$\mathcal{A}$” by “$R Mod$” (if that’s what you are referring to) to make the formula consistent, because you had “$R Mod$” on the right of the map (here). We can replace by $\mathcal{A}$, but then we need to do it consistently throughout.
I used $\mathcal{A}$ with an eye towards rewriting the two articles for a general such category instead of $R$Mod, and since it was already used elsewhere in the article. But I didn’t want to do that rewrite, because firstly I only feel everything should work out and don’t really have the time to do the calculations right now and second because of the use of the direct sum which was a new notion to me. I thought it should just be the coproduct and am unsure about the relevance of using one of the notions from direct sum instead.
Oh, I see, you are wondering why the entry as a whole declares that $\mathcal{A} = R Mod$. For no reason, I suppose. Looks like I made that move 11 years ago in revision 3, but we can change this.
In general, this entry could do with a thorough polishing/expansion/rewrite.
Regarding the direct sum: Yes, it’s just the finite coproduct for $\mathcal{A}$ any additive category.
Yes, exactly, that’s what I was wondering. If I find some time I’ll edit the article.
Re direct product: Makes sense, that it would agree with the finite biproducts of additive categories. I thought a bit more about relating these to the coproducts and made a comment about my findings in the discussion of direct sum. Still this leaves the infinite case open which is relevant to unbounded chain complexes. But I think in this case what we want is the coproduct, not the direct sum. Otherwise defining the differential seems problematic and we do not get the direct sum commuting with the tensor product of the base category for free.
Are you referring to the definition here?
Yes, that “direct sum” is the coproduct, I have added a parenthesis making this explicit.
Yes, exactly, thank you.
added pointer to:
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