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nLab > Latest Changes: internal hom of chain complexes

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- 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…
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>ℤ</mi></mrow><annotation encoding="application/x-tex">\mathbb{Z}</annotation></semantics></math>$ -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.)
- CommentRowNumber3.
- CommentAuthorUrs
- CommentTimeAug 26th 2012
- PermaLink
Author: Urs
Format: MarkdownItexSure. I have changed it.

Sure. I have changed it.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>X</mi></mrow><annotation encoding="application/x-tex">X</annotation></semantics></math>$ and $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Y</mi></mrow><annotation encoding="application/x-tex">Y</annotation></semantics></math>$ are non-negatively graded, or non-positively graded, then $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">[</mo><mi>X</mi><mo>,</mo><mi>Y</mi><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">[X,Y]</annotation></semantics></math>$ 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?
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">[</mo><mi>X</mi><mo>,</mo><mi>Y</mi><msub><mo stretchy="false">]</mo> <mi>n</mi></msub><mo>=</mo><msub><mo lspace="0.16667em" rspace="0.16667em">∏</mo> <mrow><mi>p</mi><mo>≥</mo><mn>0</mn></mrow></msub><mi>Hom</mi><mo stretchy="false">(</mo><msub><mi>X</mi> <mi>p</mi></msub><mo>,</mo><msub><mi>Y</mi> <mrow><mi>n</mi><mo>+</mo><mi>p</mi></mrow></msub><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">[X, Y]_n = \prod_{p \geq 0} Hom(X_p, Y_{n+p})</annotation></semantics></math>$ ?
- 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” $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>U</mi></mrow><annotation encoding="application/x-tex">U</annotation></semantics></math>$ of non-negatively-graded cochain complexes into unbounded complexes is fully faithful, strongly monoidal, and has a right adjoint, say $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>R</mi></mrow><annotation encoding="application/x-tex">R</annotation></semantics></math>$ , so
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>Hom</mi><mo stretchy="false">(</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo>⊗</mo><mi>B</mi><mo stretchy="false">)</mo><mo>,</mo><mi>C</mi><mo stretchy="false">)</mo><mo>≅</mo><mi>Hom</mi><mo stretchy="false">(</mo><mi>U</mi><mi>A</mi><mo>⊗</mo><mi>U</mi><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">)</mo><mo>≅</mo><mi>Hom</mi><mo stretchy="false">(</mo><mi>U</mi><mi>A</mi><mo>,</mo><mo stretchy="false">[</mo><mi>U</mi><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">]</mo><mo stretchy="false">)</mo><mo>≅</mo><mi>Hom</mi><mo stretchy="false">(</mo><mi>A</mi><mo>,</mo><mi>R</mi><mo stretchy="false">[</mo><mi>U</mi><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">]</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">Hom (U (A \otimes B), C) \cong Hom (U A \otimes U B, C) \cong Hom (U A, [U B, C]) \cong Hom (A, R [U B, C])</annotation></semantics></math>$
therefore non-negatively-graded cochain complexes is also monoidal closed. The functor $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>R</mi></mrow><annotation encoding="application/x-tex">R</annotation></semantics></math>$ 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…
- 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.
- 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?)
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi><mi>Ω</mi><mo>=</mo><mi>Id</mi></mrow><annotation encoding="application/x-tex">\Sigma\Omega=Id</annotation></semantics></math>$ . 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo>⊗</mo></mrow><annotation encoding="application/x-tex">\otimes</annotation></semantics></math>$ is cocontinuous in each variable, $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi><mi>Ω</mi><mo>=</mo><mi>Id</mi></mrow><annotation encoding="application/x-tex">\Sigma\Omega=Id</annotation></semantics></math>$ gives $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi><mi>I</mi><mo>⊗</mo><mi>Ω</mi><mi>I</mi><mo>≅</mo><mi>I</mi></mrow><annotation encoding="application/x-tex">\Sigma I \otimes \Omega I \cong I</annotation></semantics></math>$ , where $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>I</mi></mrow><annotation encoding="application/x-tex">I</annotation></semantics></math>$ is the monoidal unit, so the functor $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi><mo>≅</mo><mo stretchy="false">(</mo><mi>Σ</mi><mi>I</mi><mo>⊗</mo><mo lspace="0.11111em" rspace="0em">−</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\Sigma \cong (\Sigma I \otimes -)</annotation></semantics></math>$ is an equivalence and we are stable.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo>•</mo><mo>→</mo><mo>•</mo><mo>→</mo><mo>•</mo><mo>→</mo><mi>⋯</mi></mrow><annotation encoding="application/x-tex">\bullet \to \bullet \to \bullet \to \cdots</annotation></semantics></math>$ instead of a diagram of shape $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>⋯</mi><mo>→</mo><mo>•</mo><mo>→</mo><mo>•</mo><mo>→</mo><mi>⋯</mi></mrow><annotation encoding="application/x-tex">\cdots \to \bullet \to \bullet \to \cdots</annotation></semantics></math>$ . But $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>U</mi></mrow><annotation encoding="application/x-tex">U</annotation></semantics></math>$ isn’t induced by a functor between diagram shapes, unfortunately.

I guess you are assuming that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi></mrow><annotation encoding="application/x-tex">\Sigma</annotation></semantics></math>$ and $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Ω</mi></mrow><annotation encoding="application/x-tex">\Omega</annotation></semantics></math>$ are strong functors or something like that? Otherwise I do not see why $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi><mi>I</mi><mo>⊗</mo><mi>Ω</mi><mi>I</mi><mo>≅</mo><mi>I</mi></mrow><annotation encoding="application/x-tex">\Sigma I \otimes \Omega I \cong I</annotation></semantics></math>$ . But then why should they be strong functors?
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>U</mi></mrow><annotation encoding="application/x-tex">U</annotation></semantics></math>$ is induced by a functor of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>Set</mi> <mo>*</mo></msub></mrow><annotation encoding="application/x-tex">Set_*</annotation></semantics></math>$ -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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>Set</mi> <mo>*</mo></msub></mrow><annotation encoding="application/x-tex">Set_*</annotation></semantics></math>$ -Cauchy-complete them), then can’t we just send all the positive objects to $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mn>0</mn></mrow><annotation encoding="application/x-tex">0</annotation></semantics></math>$ ?

$<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi></mrow><annotation encoding="application/x-tex">\Sigma</annotation></semantics></math>$ is a colimit construction, so it’s preserved by $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo>⊗</mo></mrow><annotation encoding="application/x-tex">\otimes</annotation></semantics></math>$ , so we get
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>Σ</mi><mi>I</mi><mo>⊗</mo><mi>Ω</mi><mi>I</mi><mo>≅</mo><mi>Σ</mi><mo stretchy="false">(</mo><mi>I</mi><mo>⊗</mo><mi>Ω</mi><mi>I</mi><mo stretchy="false">)</mo><mo>≅</mo><mi>Σ</mi><mi>Ω</mi><mi>I</mi><mo>≅</mo><mi>I</mi></mrow><annotation encoding="application/x-tex">\Sigma I \otimes \Omega I \cong \Sigma(I\otimes \Omega I) \cong \Sigma\Omega I \cong I</annotation></semantics></math>$
the last step by the assumption that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Σ</mi><mi>Ω</mi><mo>=</mo><mi>Id</mi></mrow><annotation encoding="application/x-tex">\Sigma\Omega=Id</annotation></semantics></math>$ .
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>U</mi></mrow><annotation encoding="application/x-tex">U</annotation></semantics></math>$ 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>A</mi> <mo>•</mo></msup></mrow><annotation encoding="application/x-tex">A^{\bullet}</annotation></semantics></math>$ to $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>coker</mi><mo stretchy="false">(</mo><msup><mi>A</mi> <mrow><mo lspace="0.11111em" rspace="0em">−</mo><mn>1</mn></mrow></msup><mo>→</mo><msup><mi>A</mi> <mn>0</mn></msup><mo stretchy="false">)</mo><mo>→</mo><msup><mi>A</mi> <mn>1</mn></msup><mo>→</mo><msup><mi>A</mi> <mn>2</mn></msup><mo>→</mo><mi>⋯</mi></mrow><annotation encoding="application/x-tex">coker (A^{-1} \to A^0) \to A^1 \to A^2 \to \cdots</annotation></semantics></math>$ . I guess that resolves the duality question.
- CommentRowNumber13.
- CommentAuthorMike Shulman
- CommentTimeJun 9th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexYes, that makes sense.

Yes, that makes sense.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>𝒞𝒽</mi></mrow><annotation encoding="application/x-tex">\mathcal{Ch}</annotation></semantics></math>$ 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
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>𝒞𝒽</mi><mo stretchy="false">(</mo><mi>A</mi><mo>⊗</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">)</mo><mo>≅</mo><mi>𝒞𝒽</mi><mo stretchy="false">(</mo><mi>A</mi><mo>,</mo><mo stretchy="false">[</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">]</mo><mo stretchy="false">)</mo><mo>,</mo></mrow><annotation encoding="application/x-tex"> \mathcal{Ch}(A\otimes B,C) \cong \mathcal{Ch}(A, [B,C]), </annotation></semantics></math>$
but instead
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mo stretchy="false">[</mo><mi>A</mi><mo>⊗</mo><mi>B</mi><mo>,</mo><mi>C</mi><msub><mo stretchy="false">]</mo> <mn>0</mn></msub><mo>≅</mo><mo stretchy="false">[</mo><mi>A</mi><mo>,</mo><mo stretchy="false">[</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">]</mo><msub><mo stretchy="false">]</mo> <mn>0</mn></msub><mo>.</mo></mrow><annotation encoding="application/x-tex"> [A\otimes B, C]_0 \cong [A,[B,C]]_0. </annotation></semantics></math>$
The former failing is most easily seen when taking $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><mo>=</mo><mi>I</mi></mrow><annotation encoding="application/x-tex">A=I</annotation></semantics></math>$ , the unit of the tensor product.

However this category should be more useful than a potential closed structure on $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>𝒞𝒽</mi></mrow><annotation encoding="application/x-tex">\mathcal{Ch}</annotation></semantics></math>$ as the internal hom in this case would not include homotopies, since it only contains chain maps.
- 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.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><mo>=</mo><mi>I</mi></mrow><annotation encoding="application/x-tex">A = I</annotation></semantics></math>$ the unit of the tensor product of chain complexes, then
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>𝒞𝒽</mi><mo stretchy="false">(</mo><mi>I</mi><mo>⊗</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">)</mo><mo>≅</mo><mi>𝒞𝒽</mi><mo stretchy="false">(</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">)</mo><mo>≠</mo><mo stretchy="false">[</mo><mi>B</mi><mo>,</mo><mi>C</mi><msub><mo stretchy="false">]</mo> <mn>0</mn></msub><mo>≅</mo><mi>𝒞𝒽</mi><mo stretchy="false">(</mo><mi>I</mi><mo>,</mo><mo stretchy="false">[</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo stretchy="false">]</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex"> \mathcal{Ch}(I\otimes B,C)\cong\mathcal{Ch}(B,C) \neq [B,C]_0 \cong \mathcal{Ch}(I,[B,C]) </annotation></semantics></math>$
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.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>I</mi></mrow><annotation encoding="application/x-tex">I</annotation></semantics></math>$ 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.]
- 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.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>Ch</mi><mo stretchy="false">(</mo><mi>I</mi><mo>,</mo><mo lspace="0.11111em" rspace="0em">−</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">Ch(I,-)</annotation></semantics></math>$ .

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

diff, v4, current
- CommentRowNumber20.
- CommentAuthorUrs
- CommentTimeJul 5th 2023
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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
- CommentRowNumber21.
- CommentAuthorjulia9367
- CommentTimeJul 5th 2023
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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.
- CommentRowNumber22.
- CommentAuthorUrs
- CommentTimeAug 23rd 2023
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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 [[doi:10.1007/978-3-642-99902-4](https://doi.org/10.1007/978-3-642-99902-4)] <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

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