At least twice on Math Stack Exchange, the question has been asked, "Why is the tensor product of two chain complexes defined the way it is?" See for instance here and here. And yet I've still somehow managed to fail to understand either answer.

That's not to say that the answers haven't given me some greater insight into what is going on. They did teach me that I should regard a chain complex of modules as a graded module with a very specific nilpotent endomorphism associated to it. And that made me realize "Oh, so that's why you end up with that particular grading, the direct sum of etc., etc." But the specific definition of the differential still baffled me.

But, I'm a stubborn bastard, so I figure, "Well, I know the definition of tensor products of groups, it's in terms of universal properties! Maybe there is something similar here I can go for? Maybe there is such a thing as a general definition of a tensor product for a category, generalized from the definition of the one for groups and modules, and once I've understood that, then it's merely going to be a matter of applying that definition for the specific category of chain complexes of modules, and everything will make sense!"

And what do you know, nLab had the following to say:

For $M$ a multicategory and $A$ and $B$ objects in $M$, the tensor product $A \otimes B$ is defined to be an object equipped with a universal multimorphism $A,B\to A \otimes B$ in that any multimorphism $A,B\to C$ factors uniquely through $A,B\to A \otimes B$ via a (1-ary) morphism $A \otimes B\to C$.

Well, that's great, I figure! Now I just need to start by constructing a multimorphism of chain complexes, and-...

And then it hits me that I don't know how to define that either.


Googling doesn't seem to lead me anywhere useful, so I'm asking the community, how does one define a multimorphism of chain complexes of modules/groups?

Looking forward to your answers!

  • 2
    $\begingroup$ A multimorphism $(A,B)\to C$ is, of course, just a morphism $A\otimes B\to C.$ :) Seriously, though, this perspective isn’t going to get you anywhere better. There’s also good discussion on MO here: mathoverflow.net/questions/359306/… But at the end of the day, I’m pretty sure the real reason why we define the tensor product of chain complexes the way we do is because that’s how you get the singular complex of a product of topological spaces. $\endgroup$ Commented Jul 30, 2022 at 21:28

2 Answers 2


I don't recommend thinking about things this way; this can be done but I don't think it's particularly enlightening. Instead here is a general question: what sorts of objects should we expect to have an operation that deserves to be called a "tensor product"?

In my opinion it's hard to do better than "any objects that look like abelian groups or sheaves of abelian groups." This is a very general idea that includes modules over commutative rings (these are quasicoherent sheaves over affine schemes), representations of groups (these are quasicoherent sheaves over stacks), and also various classes of topological vector spaces. Tensor products here then come from bilinearity, suitably interpreted.

And it includes chain complexes! This is because, by the Dold-Kan correspondence, (non-negatively graded) chain complexes (of abelian groups) are equivalent to simplicial abelian groups, so they are an alternative way to describe abelian group objects in the category of simplicial sets. This is a model-dependent version of a model-independent statement, which is that chain complexes (of abelian groups) describe (some) homotopy-theoretic objects known as spectra, which are the homotopy-theoretic version of abelian groups.

There is a homotopy-theoretic operation on spectra called the smash product which is the tensor product in this context, and the tensor product of chain complexes (sometimes) models this. This gives a (very indirect) way of explaining why the tensor product of chain complexes ought to exist and be a meaningful operation at all.

Edit: The relationship to singular chains and singular homology is that singular chains on a space $X$ models a homotopy-theoretic construction called the "free $\mathbb{Z}$-module spectrum" on $X$, which is a homotopy-theoretic version of the free abelian group on a set. The homotopy groups of this spectrum are then the homology of $X$. The Dold-Thom theorem is a concrete model-dependent version of this. Then taking products of spaces corresponds to taking smash products of these spectra which mirrors the corresponding phenomenon for taking free abelian groups and that's where the Eilenberg-Zilber theorem comes from abstractly.

  • $\begingroup$ Smash products correspond to "derived tensor products". $\endgroup$
    – Yai0Phah
    Commented Jul 31, 2022 at 0:06

$\def\ch{\operatorname{Ch}} \def\hotimes{\hat{\otimes}}$I found out about the following by playing around with the math by myself, for I couldn't find a reference online, so please, let me know if you know some :).

One can verify that applying either $\partial_{X\otimes (Y\otimes Z)}$ or $\partial_{(X\otimes Y)\otimes Z}$ to $x\otimes y\otimes z$ gives $\partial_Xx\otimes y\otimes z+(-1)^{|x|}x\otimes\partial_Yy\otimes z+(-1)^{|x|+|y|}x\otimes y\otimes \partial_Z z$. Thus $X\otimes (Y\otimes Z)\cong(X\otimes Y)\otimes Z$ for chain complexes $X,Y,Z$. This suggests the following generalization: we define the left (resp., right) tensor product $A_1^\bullet\otimes\cdots\otimes A_n^\bullet$ (resp., $A_1^\bullet\hotimes\cdots\hotimes A_n^\bullet$) of the chain complexes $A_i^\bullet$ to be the complex whose chains of degree $k$ equal $\displaystyle\bigoplus_{k_1+\dots +k_n=k}A_1^{k_1}\otimes\cdots\otimes A_n^{k_n}$, and where the differential at the chain $a=a_1\otimes\cdots \otimes a_n$ is given by $$ \partial(a_1\otimes\cdots \otimes a_n)=\sum_{i=1}^n(-1)^{|a_1|+\dots+|a_{i-1}|}\partial_ia $$ (resp., $$ \hat{\partial}(a_1\otimes\cdots \otimes a_n)=\sum_{i=1}^n(-1)^{|a_{i+1}|+\dots+|a_{n}|}\partial_ia $$ ), where $\partial_ia=a_1\otimes\cdots \otimes a_{i-1}\otimes\partial a_i\otimes a_{i+1}\otimes\cdots\otimes a_n$. Note that $\hat{\partial}=\sigma\circ\partial\circ\sigma$, where $\sigma$ is the isomorphism $(a_1,\dots,a_n)\leftrightarrow(a_n,\dots,a_1)$.

Its square is zero:

\begin{align*} \partial\partial a &=\sum_{i=1}^n(-1)^{|a_1|+\cdots+|a_{i-1}|}\partial\partial_ia\\ &=\sum_{i=1}^n(-1)^{|a_1|+\cdots+|a_{i-1}|} \left( \sum_{j=1}^{i-1}(-1)^{|a_1|+\cdots+|a_{j-1}|}\partial_j\partial_ia +\sum_{j=i+1}^{n}(-1)^{|a_1|+\cdots+|a_{j-1}|+1}\partial_j\partial_ia \right)\\ &=\sum_{i=2}^n\sum_{j=1}^{i-1}(-1)^{|a_j|+\cdots+|a_{i-1}|}\partial_j\partial_i a +\sum_{i=1}^{n-1}\sum_{j=i+1}^n(-1)^{|a_i|+\dots+|a_{j-1}|+1}\partial_j\partial_i a. \end{align*} The first term equals $\displaystyle\sum_{j=1}^{n-1}\sum_{i=j+1}^n(-1)^{|a_j|+\cdots+|a_{i-1}|}\partial_j\partial_i a$, which under a change of indexes $i\leftrightarrow j$ it becomes minus the second term; thus $\partial\partial a=0$. Therefore, also $\hat{\partial}^2=(\sigma\circ\partial\circ\sigma)\circ(\sigma\circ\partial\circ\sigma)=\sigma\circ\partial\circ\partial\circ\sigma=0$.

Equation $\hat{\partial}\circ\sigma=\sigma\circ\partial$ means that $\sigma$ is an isomorphism $A_1^\bullet\otimes\cdots \otimes A_n^\bullet\cong A_n^\bullet\hotimes\cdots\hotimes A_1^\bullet$ of chain complexes.

Definition. A left (resp., right) multimorphism of cochain complexes $A_1^\bullet\times\cdots\times A_n^\bullet\to B^\bullet$ is a collection of multilinear maps $\varphi=\varphi_{k_1\dots k_n}:A_1^{k_1}\times\cdots\times A_n^{k_n}\to B^{k_1+\dots+k_n}$, $(k_1,\dots,k_n)\in\mathbb{Z}^n$ such that $$ \partial_B(\varphi(x))=\sum_{i=1}^n(-1)^{|x_1|+\cdots+|x_{i-1}|}\varphi(\partial_ix) $$ (resp., $$ \partial_B(\varphi(x))=\sum_{i=1}^n(-1)^{|x_{i+1}|+\cdots+|x_n|}\varphi(\partial_ix)) $$ where $x=(x_1,\dots,x_n)$, $x_j\in A_j^{k_j}$, $|x_j|=k_j$, and we write $\partial_ix=(x_1,\dots,x_{i-1},\partial_{A_i}x_i,x_{i+1},\dots,x_n)$.

Note that for $n=1$, a left or right multimorphism of cochain complexes is the same as a cochain map between two complexes. Given a collection $\varphi=(\varphi_{k_1\dots k_n}:A_1^{k_1}\times\cdots\times A_n^{k_n}\to B^{k_1+\dots+k_n})$ of multilinear maps, we have that $\varphi$ is a left (resp., right) multimorphism if and only if $\hat{\varphi}=(\hat{\varphi}_{k_1\dots k_n})$ is a right (resp., left) multimorphism, where $\hat{\varphi}_{k_1\dots k_n}=\varphi_{k_n\dots k_1}\circ \sigma_{k_1\dots k_n}$, where $\sigma_{k_1\dots k_n}$ is the canonical bijection $A_1^{k_1}\times\cdots\times A_n^{k_n}\cong A_n^{k_n}\times\cdots\times A_1^{k_1}$.

Denote $\ch(A_1,\dots,A_n;B)$ (resp., $\widehat{\ch}(A_1,\dots,A_n;B)$) to the set of all left (resp., right) multimorphisms $A_1^\bullet\times\cdots\times A_n^\bullet\to B^\bullet$. Multicomposition (see def. 2.1.1 here) is given by $$ \ch(B_1,\dots,B_n;C)\times \ch(A_{11},\dots,A_{1r_1};B_1)\times\cdots\times \ch(A_{n1},\dots,A_{nr_n};B_n)\\\longrightarrow \ch(A_{11},\dots,A_{1r_1},\dots,A_{n1},\dots,A_{nr_n};C)\\ ((\psi_{k_1\dots k_n}),(\varphi^1_{k_{11}\dots k_{1r_1}}),\dots,(\varphi^n_{k_{n1}\dots k_{nr_n}}))\mapsto(\psi_{k_1\dots k_n}(\varphi^1_{k_{11}\dots k_{1r_1}},\dots,\varphi^n_{k_{n1}\dots k_{nr_n}})), $$ where the last collection is over all posible values $(k_{11},\dots,k_{1r_1},\dots,k_{n1},\dots,k_{nr_n})\in\mathbb{Z}^{r_1+\dots+r_n}$ and $k_i=k_{i1}+\dots+k_{ir_i}$. One can verify that the multicomposite is indeed a left multimorphism. The formula for composition of right multimorphisms is the same but one changes $\ch$ by $\widehat{\ch}$ above.

Note that the canonical map $A_1\times\cdots\times A_n\to A_1\otimes\cdots\otimes A_n$ (resp., $A_1\times\cdots\times A_n\to A_1\hotimes\cdots\hotimes A_n$) is a left (resp., right) multimorphism of cochain complexes. We arrive to the result we longed for.

Lemma. In the multicategory $\ch$ (resp., $\widehat{\ch}$) of left (resp., right) multimorphisms of cochain complexes, the tensor product (in the multicategorical sense) is the left (resp., right) tensor product of cochain complexes.

Proof. In plain terms: the statement asserts that precomposition by $A_1\times\cdots\times A_n\to A_1\otimes\cdots\otimes A_n$ induces an isomorphism $$ \ch(A_1\otimes\cdots\otimes A_n,B)\xrightarrow{\cong}\ch(A_1,\dots, A_n;B). $$ From our definitions, it follows that precomposition by $A_1\times\cdots\times A_n\to A_1\otimes\cdots\otimes A_n$ induces a one-to-one correspondence between collections of multilinear maps $(A_1^{k_1}\times\cdots\times A_n^{k_n}\to B^{k_1+\dots+k_n})_{(k_1,\dots, k_n)\in\mathbb{Z}^n}$ and the collections of linear maps $((A_1\otimes\cdots\otimes A_n)^k\to B^k)_{k\in\mathbb{Z}}$. It is left to verify that this correspondence restricts to collections of maps respecting differentials. But this is basically the definition of a left multimorphism and of the differential of the left tensor product.

Changing “$\ch$” by “$\widehat{\ch}$” above and “left” by “right” proves the analogous assertion. $\square$


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