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I am trying to make sense of the introduction to Section 2 of the following paper.

Let $R$ be a $\mathbb{Z}$-graded ring and $\text{Mod}_R$ be the 'category of graded $R$-modules'. Let $D$ be the contravariant Hom functor $\text{Hom}_R(-,R)$. I am confused about the precise details of the category $\text{Mod}_R$, which I assume must be abelian however it is defined since the authors claim that self-injectivity of $R$ implies that $D$ is exact.

I first assumed the morphisms in $\text{Mod}_R$ were the homogeneous $R$-linear maps of degree zero i.e. those for which $f(M_i)\subseteq N_i$ for all $i$. However it seems that this does not make $\text{Hom}_0(M,N)$ an $R$-module, just an abelian group.

One other possibility would be $\sum_i\text{Hom}_i(M,N)$ i.e. those with $f=\sum_if_i$, $f_i$ homogeneous of degree $i$. This is a graded module, but the kernel of such morphisms do not appear to be submodules.

What exactly is $DM$ and what is the correct understanding of $\text{Mod}_R$? And is the correct choice of $\text{Mod}_R$ an abelian category?

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  • $\begingroup$ How is it a problem that Hom(M,N) is just an abelian group? $\endgroup$
    – Eric Wofsey
    Jul 13 at 15:53
  • $\begingroup$ I suppose for general $M$ and $N$ it perhaps isn't necessary, but I certainly need $DM:=\text{Hom}_R(M,R)$ to be a graded $R$-module for each $M$. If $f$ is homogeneous of degree $0$ and $a\in R_1$, then $af$ has degree $1$, so degree zero morphisms don't seem to be a submodule with the definitions I currently have $\endgroup$
    – user829347
    Jul 13 at 15:56
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    $\begingroup$ I think the two standard choices for morphisms are either (a) all homogeneous maps of degree $0$, or (b) all homogeneous maps of any degree. You can consider case (b) as a graded object but only if you agree to never add morphisms of different gradings, because kernels of morphisms of the form $\Sigma f_i$ need not be graded. $\endgroup$
    – John Palmieri
    Jul 13 at 17:46
  • $\begingroup$ @JohnPalmieri I'm a bit confused by this. Do you have any idea what the morphisms should be in the paper I linked to? $\endgroup$
    – user829347
    Jul 13 at 18:22
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    $\begingroup$ It might be option (b), but as @EricWofsey says in their answer, you should think of graded objects as sequences of objects, not direct sums. Does it actually matter what the morphisms are? If you take the viewpoint of Eric's answer, then $DM$ is defined regardless. At a quick glance, it appears that the authors use degree $0$ maps in other parts of the paper, so maybe the answer is (a). $\endgroup$
    – John Palmieri
    Jul 13 at 19:42

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The morphisms in the category $\text{Mod}_R$ are just the degree $0$ homomorphisms. However, in this context, $\operatorname{Hom}_R(M,R)$ is referring not to the Hom-set of the category $\text{Mod}_R$, but to the internal hom of graded objects, which is the graded $R$-module whose degree $i$ part is the degree $i$ homomorphisms.

Consequently, it does not just follow immediately from the definition that $D$ is exact if $R$ is self-injective, since self-injectivity is about only the degree $0$ homomorphisms. However, it follows easily since degree $i$ homomorphisms $M\to R$ are the same thing as homomorphisms $\Sigma^{-i}M\to R$, so self-injectivity of $R$ implies that the degree $i$ part of $DM$ is an exact functor $\text{Mod}_R\to\text{Ab}$ for each $i$ and thus $D$ itself is exact.

(A good way to think about this is that $\text{Mod}_R$ is not just an ordinary category but a category enriched in graded abelian groups, with the enriched Hom being the graded abelian group whose degree $i$ part is the degree $i$ homomorphisms. The forgetful functor to sets which turns this enriched category back into an ordinary category does not take a graded abelian group to the direct sum of its parts of each degree, but instead takes it to only the degree $0$ part.

Relatedly, in this context, you should always think of graded objects not as direct sums but simply as sequences, so a graded abelian group $A$ is not an abelian group with a direct sum decomposition $A=\bigoplus A_n$ but is simply a sequence $(A_n)$ of abelian groups. This is virtually always the right way to think about graded objects in the context of algebraic topology and homological algebra.)

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  • $\begingroup$ Thanks for this. In your final paragraph, do you mean thinking of them as completely separate objects with connecting maps $A_i\times A_j\to A_{i+j}$ and then show that this forms a ring/module etc? Surely it must be necessary to consider the whole thing as one structure for some purposes.. almost everywhere I look it is defined as a decomposition but I see your point $\endgroup$
    – user829347
    Jul 13 at 23:54
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    $\begingroup$ Right, they are completely separate objects. If you are just talking about a graded abelian group, then there is no additional structure relating them. If you are talking about a graded ring, for instance, then you also have multiplication maps $A_i\times A_j\to A_{i+j}$ (which must satisfy a bunch of axioms). $\endgroup$
    – Eric Wofsey
    Jul 14 at 0:21
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    $\begingroup$ For the sorts of graded objects you care about in algebraic topology, there really is no need to ever form the direct sum and think of them as a single set rather than as a sequence. You just don't care about sums of elements in different degrees. (And some things just don't work the way you want if you try to consider the direct sum as a single algebraic object; for instance this.) $\endgroup$
    – Eric Wofsey
    Jul 14 at 0:24
  • $\begingroup$ Thanks again. Do you know a good reference for this approach to dealing with gradings? $\endgroup$
    – user829347
    Jul 14 at 0:34
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    $\begingroup$ No, but everything works out straightforwardly--there aren't really any surprises to be aware of. $\endgroup$
    – Eric Wofsey
    Jul 14 at 0:58

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