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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeAug 26th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexI started putting some content into _[[internal hom of chain complexes]]_. But am being interrupted now...

   I started putting some content into internal hom of chain complexes. But am being interrupted now…

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 26th 2012
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexAre these non-negatively graded chain complexes? Because if you want $\mathbb{Z}$-graded, then you need to use a product (not a sum) in the construction. (Basically, I think it would be clearer anyway to use a product, even for non-negatively graded objects.)

   Are these non-negatively graded chain complexes? Because if you want $\mathbb{Z}$-graded, then you need to use a product (not a sum) in the construction. (Basically, I think it would be clearer anyway to use a product, even for non-negatively graded objects.)

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeAug 26th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexSure. I have changed it.

   Sure. I have changed it.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeJun 7th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThe definition of the internal-hom of chain complexes doesn't respect non-negative or non-positive grading. That is, even if $X$ and $Y$ are non-negatively graded, or non-positively graded, then $[X,Y]$ may not be either. Can one define a variant internal-hom that applies in either of these situations? Or are we just forced into the unbounded case if we want internal-homs?

   The definition of the internal-hom of chain complexes doesn’t respect non-negative or non-positive grading. That is, even if $X$ and $Y$ are non-negatively graded, or non-positively graded, then $[X,Y]$ may not be either. Can one define a variant internal-hom that applies in either of these situations? Or are we just forced into the unbounded case if we want internal-homs?

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 7th 2014
   • (edited Jun 7th 2014)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexFor the positively graded case, can't we just use $[X, Y]_n = \prod_{p \geq 0} Hom(X_p, Y_{n+p})$?

   For the positively graded case, can’t we just use $[X, Y]_n = \prod_{p \geq 0} Hom(X_p, Y_{n+p})$?

6. • CommentRowNumber6.
   • CommentAuthorZhen Lin
   • CommentTimeJun 8th 2014
   • (edited Jun 8th 2014)
   • PermaLink
   Author: Zhen Lin
   Format: MarkdownItexThe "inclusion" $U$ of non-negatively-graded cochain complexes into unbounded complexes is fully faithful, strongly monoidal, and has a right adjoint, say $R$, so $$Hom (U (A \otimes B), C) \cong Hom (U A \otimes U B, C) \cong Hom (U A, [U B, C]) \cong Hom (A, R [U B, C])$$ therefore non-negatively-graded cochain complexes is also monoidal closed. The functor $R$ is the good truncation, which replaces negative degrees with the zero object and the object in degree 0 with the kernel of the differential. I suppose the story for chain complexes has a _left_ adjoint instead, which is troublesome...

   The “inclusion” $U$ of non-negatively-graded cochain complexes into unbounded complexes is fully faithful, strongly monoidal, and has a right adjoint, say $R$, so

   $$Hom (U (A \otimes B), C) \cong Hom (U A \otimes U B, C) \cong Hom (U A, [U B, C]) \cong Hom (A, R [U B, C])$$

   therefore non-negatively-graded cochain complexes is also monoidal closed. The functor $R$ is the good truncation, which replaces negative degrees with the zero object and the object in degree 0 with the kernel of the differential. I suppose the story for chain complexes has a left adjoint instead, which is troublesome…

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeJun 8th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThat's about the conclusion I had come to as well. It's curious that the two cases seem so different.

   That’s about the conclusion I had come to as well. It’s curious that the two cases seem so different.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJun 8th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex(@Zhen, why did you put "inclusion" in quotes?)

   (@Zhen, why did you put “inclusion” in quotes?)

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeJun 8th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI asked because I was wondering whether you could have a closed monoidal homotopy theory which is not stable, but where $\Sigma\Omega=Id$. For a while I thought that non-positively graded chain complexes should be an answer, but then I ran into this problem. Now I think I have a proof that this is impossible: since $\otimes$ is cocontinuous in each variable, $\Sigma\Omega=Id$ gives $\Sigma I \otimes \Omega I \cong I$, where $I$ is the monoidal unit, so the functor $\Sigma \cong (\Sigma I \otimes -)$ is an equivalence and we are stable.

   I asked because I was wondering whether you could have a closed monoidal homotopy theory which is not stable, but where $\Sigma\Omega=Id$. For a while I thought that non-positively graded chain complexes should be an answer, but then I ran into this problem. Now I think I have a proof that this is impossible: since $\otimes$ is cocontinuous in each variable, $\Sigma\Omega=Id$ gives $\Sigma I \otimes \Omega I \cong I$, where $I$ is the monoidal unit, so the functor $\Sigma \cong (\Sigma I \otimes -)$ is an equivalence and we are stable.

10. • CommentRowNumber10.
    • CommentAuthorZhen Lin
    • CommentTimeJun 8th 2014
    • PermaLink
    Author: Zhen Lin
    Format: MarkdownItexOh, I suppose because I was thinking of it as being a diagram of shape $\bullet \to \bullet \to \bullet \to \cdots$ instead of a diagram of shape $\cdots \to \bullet \to \bullet \to \cdots$. But $U$ isn't induced by a functor between diagram shapes, unfortunately. I guess you are assuming that $\Sigma$ and $\Omega$ are strong functors or something like that? Otherwise I do not see why $\Sigma I \otimes \Omega I \cong I$. But then why should they be strong functors?

    Oh, I suppose because I was thinking of it as being a diagram of shape $\bullet \to \bullet \to \bullet \to \cdots$ instead of a diagram of shape $\cdots \to \bullet \to \bullet \to \cdots$. But $U$ isn’t induced by a functor between diagram shapes, unfortunately.

    I guess you are assuming that $\Sigma$ and $\Omega$ are strong functors or something like that? Otherwise I do not see why $\Sigma I \otimes \Omega I \cong I$. But then why should they be strong functors?

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeJun 8th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexI would have thought that $U$ is induced by a functor of $Set_*$-enriched diagrams, which is where we have to be to consider chain complexes as diagrams anyway. If we include zero objects in our diagrams (i.e. we $Set_*$-Cauchy-complete them), then can't we just send all the positive objects to $0$? $\Sigma$ is a colimit construction, so it's preserved by $\otimes$, so we get $$\Sigma I \otimes \Omega I \cong \Sigma(I\otimes \Omega I) \cong \Sigma\Omega I \cong I$$ the last step by the assumption that $\Sigma\Omega=Id$.

    I would have thought that $U$ is induced by a functor of $Set_*$-enriched diagrams, which is where we have to be to consider chain complexes as diagrams anyway. If we include zero objects in our diagrams (i.e. we $Set_*$-Cauchy-complete them), then can’t we just send all the positive objects to $0$?

    $\Sigma$ is a colimit construction, so it’s preserved by $\otimes$, so we get

    $$\Sigma I \otimes \Omega I \cong \Sigma(I\otimes \Omega I) \cong \Sigma\Omega I \cong I$$

    the last step by the assumption that $\Sigma\Omega=Id$.

12. • CommentRowNumber12.
    • CommentAuthorZhen Lin
    • CommentTimeJun 9th 2014
    • PermaLink
    Author: Zhen Lin
    Format: MarkdownItexYes, we would have to do that. But if $U$ is induced by a functor between diagram shapes, then it really should have both adjoints. I think the left adjoint is the functor that sends an unbounded cochain complex $A^{\bullet}$ to $coker (A^{-1} \to A^0) \to A^1 \to A^2 \to \cdots$. I guess that resolves the duality question.

    Yes, we would have to do that. But if $U$ is induced by a functor between diagram shapes, then it really should have both adjoints. I think the left adjoint is the functor that sends an unbounded cochain complex $A^{\bullet}$ to $coker (A^{-1} \to A^0) \to A^1 \to A^2 \to \cdots$. I guess that resolves the duality question.

13. • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeJun 9th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexYes, that makes sense.

    Yes, that makes sense.

14. • CommentRowNumber14.
    • CommentAuthorjulia9367
    • CommentTimeJul 4th 2023
    • (edited Jul 5th 2023)
    • PermaLink
    Author: julia9367
    Format: MarkdownItex_edit:_ This is wrong, see below or ignore. I think it should be noted, that this does not make $\mathcal{Ch}$ into a closed category, but rather a category of chain complexes with morphisms all graded maps, not just chain maps. I.e., we do not have $$ \mathcal{Ch}(A\otimes B,C) \cong \mathcal{Ch}(A, [B,C]), $$ but instead $$ [A\otimes B, C]_0 \cong [A,[B,C]]_0. $$ The former failing is most easily seen when taking $A=I$, the unit of the tensor product. However this category should be more useful than a potential closed structure on $\mathcal{Ch}$ as the internal hom in this case would not include homotopies, since it only contains chain maps.

    edit: This is wrong, see below or ignore.

    I think it should be noted, that this does not make $\mathcal{Ch}$ into a closed category, but rather a category of chain complexes with morphisms all graded maps, not just chain maps. I.e., we do not have

    $$\mathcal{Ch}(A\otimes B,C) \cong \mathcal{Ch}(A, [B,C]),$$

    but instead

    $$[A\otimes B, C]_0 \cong [A,[B,C]]_0.$$

    The former failing is most easily seen when taking $A=I$, the unit of the tensor product.

    However this category should be more useful than a potential closed structure on $\mathcal{Ch}$ as the internal hom in this case would not include homotopies, since it only contains chain maps.

15. • CommentRowNumber15.
    • CommentAuthorDmitri Pavlov
    • CommentTimeJul 4th 2023
    • PermaLink
    Author: Dmitri Pavlov
    Format: MarkdownItexRe #14: Given that Ch(X,Y) is simply the closed elements of degree 0 in [X,Y], i.e., closed elements in [X,Y]_0, the isomorphism [A⊗B,C]≅[A,[B,C]] implies both displayed formulas in #14.

    Re #14: Given that Ch(X,Y) is simply the closed elements of degree 0 in [X,Y], i.e., closed elements in [X,Y]_0, the isomorphism [A⊗B,C]≅[A,[B,C]] implies both displayed formulas in #14.

16. • CommentRowNumber16.
    • CommentAuthorjulia9367
    • CommentTimeJul 5th 2023
    • (edited Jul 5th 2023)
    • PermaLink
    Author: julia9367
    Format: MarkdownItexTake $A = I$ the unit of the tensor product of chain complexes, then $$ \mathcal{Ch}(I\otimes B,C)\cong\mathcal{Ch}(B,C) \neq [B,C]_0 \cong \mathcal{Ch}(I,[B,C]) $$ Though the third isomorphism is wrong, as Urs wrote below. I noticed this shortly after posting myself and edited it away, but put back what I originally wrote so the discussion remains readable.

    Take $A = I$ the unit of the tensor product of chain complexes, then

    $$\mathcal{Ch}(I\otimes B,C)\cong\mathcal{Ch}(B,C) \neq [B,C]_0 \cong \mathcal{Ch}(I,[B,C])$$

    Though the third isomorphism is wrong, as Urs wrote below. I noticed this shortly after posting myself and edited it away, but put back what I originally wrote so the discussion remains readable.

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeJul 5th 2023
    • (edited Jul 5th 2023)
    • PermaLink
    Author: Urs
    Format: MarkdownItexNo, the chain maps out of $I$ are the *closed* elements of degree=0 in the codomain. [edit: This was in reply to an earlier version of comment [#16](#Comment_111102), which was subsequently edited away.] [edit: Now it's back with commentary.]

    No, the chain maps out of $I$ are the closed elements of degree=0 in the codomain.

    [edit: This was in reply to an earlier version of comment #16, which was subsequently edited away.]

    [edit: Now it’s back with commentary.]

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeJul 5th 2023
    • PermaLink
    Author: Urs
    Format: MarkdownItexIf you want to do the community and the next generation a favor, then you could add a remark/example to [[internal hom of chain complexes|the entry]] to help the next reader who runs into this issue.

    If you want to do the community and the next generation a favor, then you could add a remark/example to the entry to help the next reader who runs into this issue.

19. • CommentRowNumber19.
    • CommentAuthorjulia9367
    • CommentTimeJul 5th 2023
    • PermaLink
    Author: julia9367
    Format: MarkdownItexAdds a paragraph refrasing the characterisation if the 0-cycles in terms of the canonical forgetful functor $Ch(I,-)$. See above discussion starting at [#14](https://nforum.ncatlab.org/discussion/4089/internal-hom-of-chain-complexes/?Focus=111089#Comment_111089) and the corresponding [edit](https://nforum.ncatlab.org/discussion/4261/tensor-product-of-chain-complexes/?Focus=111115#Comment_111115) at [[tensor product of chain complexes]]. <a href="https://ncatlab.org/nlab/revision/diff/internal+hom+of+chain+complexes/4">diff</a>, <a href="https://ncatlab.org/nlab/revision/internal+hom+of+chain+complexes/4">v4</a>, <a href="https://ncatlab.org/nlab/show/internal+hom+of+chain+complexes">current</a>

    Adds a paragraph refrasing the characterisation if the 0-cycles in terms of the canonical forgetful functor $Ch(I,-)$.

    See above discussion starting at #14 and the corresponding edit at tensor product of chain complexes.

    diff, v4, current

20. • CommentRowNumber20.
    • CommentAuthorUrs
    • CommentTimeJul 5th 2023
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks. Looks good. I have just made some minor adjustments ([here](https://ncatlab.org/nlab/show/internal+hom+of+chain+complexes#Properties)) to hyperlinking and formatting <a href="https://ncatlab.org/nlab/revision/diff/internal+hom+of+chain+complexes/5">diff</a>, <a href="https://ncatlab.org/nlab/revision/internal+hom+of+chain+complexes/5">v5</a>, <a href="https://ncatlab.org/nlab/show/internal+hom+of+chain+complexes">current</a>

    Thanks. Looks good.

    I have just made some minor adjustments (here) to hyperlinking and formatting

    diff, v5, current

21. • CommentRowNumber21.
    • CommentAuthorjulia9367
    • CommentTimeJul 5th 2023
    • PermaLink
    Author: julia9367
    Format: MarkdownItexThank you, also for taking the time to sort out my confusion.

    Thank you, also for taking the time to sort out my confusion.

22. • CommentRowNumber22.
    • CommentAuthorUrs
    • CommentTimeAug 23rd 2023
    • PermaLink
    Author: Urs
    Format: MarkdownItexadded pointer to: * [[Samuel Eilenberg]], [[G. Max Kelly]], §IV.6 of: *Closed Categories*, in: *[[Proceedings of the Conference on Categorical Algebra - La Jolla 1965]]*, Springer (1966) 421-562 &lbrack;[doi:10.1007/978-3-642-99902-4](https://doi.org/10.1007/978-3-642-99902-4)&rbrack; <a href="https://ncatlab.org/nlab/revision/diff/internal+hom+of+chain+complexes/6">diff</a>, <a href="https://ncatlab.org/nlab/revision/internal+hom+of+chain+complexes/6">v6</a>, <a href="https://ncatlab.org/nlab/show/internal+hom+of+chain+complexes">current</a>

    added pointer to:

    • Samuel Eilenberg, G. Max Kelly, §IV.6 of: Closed Categories, in: Proceedings of the Conference on Categorical Algebra - La Jolla 1965, Springer (1966) 421-562 [doi:10.1007/978-3-642-99902-4]

    diff, v6, current