# Formulation of Künneth theorems (definition of $\mathrm{Hom}$ and $\otimes$ of complexes)

In Rotman's An Introduction to Homological Algebra, there is written:   Questions: Let $\mathbf{A}$ and $\mathbf{A'}$ be chain complexes with differentials $\partial$ and $\partial'$ respectively.

(1) How is $\mathrm{Hom}_R(\mathbf{A},\mathbf{A}')$ defined? Is this a chain or cochain complex?

From this post, I'm guessing the solution should be similar to: $$\mathrm{Hom}_R(\mathbf{A},\mathbf{A}'):=(\prod_{i-j=n} \mathrm{Hom}_R(A_i,A'_j))_{n\in\mathbb{Z}},$$ and the differential $d\!: \prod_{i-j=n} \mathrm{Hom}_R(A_i,A'_j) \longrightarrow \prod_{i-j=n-1} \mathrm{Hom}_R(A_i,A'_j)$ is, for $\alpha_{i,j}\in\mathrm{Hom}_R(A_i,C_j)$ with $i\!-\!j\!=\!n$ and $a\!\in\!A_i$, given by $$(d\alpha_{i,j})(a):= \partial'(\alpha_{i,j}(a))-(-1)^n\alpha_{i,j}(\partial a).$$ But this does not make sense: $\partial a\in A_{i-1}$, where $\alpha_{i,j}$ is not defined, and furthermore, $\partial'(\alpha_{i,j}(a))\in A'_{j-1}$, but $i\!-\!(j\!-\!1)\neq n\!-\!1$. Furthermore, why do we in the theorem take $H^n(\mathrm{Hom}_R(\mathbf{A},\mathbf{A}'))$ instead of $H_n(\mathrm{Hom}_R(\mathbf{A},\mathbf{A}'))$?

(2) Why is the tensor product of complexes defined as $$\mathbf{A}\!\otimes\!\mathbf{A}':= (\bigoplus_{i+j=n}\!A_i\!\otimes\!A'_j)_{n\in\mathbb{Z}},$$ $$\partial\!\otimes\!\partial'(a\!\otimes\!a')\!:=\! (\partial a)\!\otimes\!a'+(-1)^{\deg a}a\!\otimes\!(\partial'\!a')?$$ I know that this is appropriate for cellular chain complexes, since an $n$-cell in a product of CW-complexes is a product of an $i$-cell and a $j$-cell with $i\!+\!j\!=\!n$. But cellular homology is just one example of many homology theories. Is there any other reason why this definition is preferred instead of other possible ones?

(3) Why is $\mathrm{Hom}_R(\mathbf{A},\mathbf{A}')$ defined the way it is, i.e. what are examples where this occurs?

(4) Must the complexes $\mathbf{A}$ and $\mathbf{A'}$ be positive for this theorem to hold? Do we get short exact sequences for all $n\!\in\!\mathbb{Z}$?

(5) Can the sequence from the last theorem be rewritten as $$0\longleftarrow \!\!\!\prod_{i+j=n}\!\!\!\mathrm{Hom}_R(H_i(\mathbf{A}),H_j(\mathbf{A}')) \longleftarrow H^n(\mathrm{Hom}_R(\mathbf{A},\mathbf{A}')) \longleftarrow \!\!\!\!\!\prod_{i+j=n-1}\!\!\! \mathrm{Ext}^1_R(H_i(\mathbf{A}),H_j(\mathbf{A}')) \longleftarrow 0?$$

• In (2), what "other possible ones" are you talking about? – user314 Mar 30 '13 at 17:04
• Oh, $A_n\!\otimes_R\!A'_n$ with $\partial\!\otimes\!\partial'(a\!\otimes\!a'):=(\partial a)\!\otimes\!(\partial'a')$, or $\bigoplus_{i+j=n}\!A_i\!\otimes\!A'_j$ with either $\partial\!\otimes\!\partial'(a\!\otimes\!a'):=(\partial a)\!\otimes\!(\partial'a')$ or $\partial\!\otimes\!\partial'(a\!\otimes\!a')\!:=\! (\partial a)\!\otimes\!a'+(-1)^{\deg a\pm\deg a'}a\!\otimes\!(\partial'\!a')$. – Leo Mar 30 '13 at 17:13
• A similar thread is this. – Leo Mar 20 '15 at 1:39

For your question (4), you don't need the complexes to be bounded -- check out section 3.6 in Weibel. Your restatement of the exact sequence in question (5) is fine too. So... let me try to address (1), (2), and (3). I'll use cohomological conventions, because that's what I'm used to and if I don't I'm sure to make an error with the indices somewhere or another. Given cochain complexes $A^\bullet$ and $B^\bullet$, the cochain complex $\mathrm{Hom}^\bullet(A^\bullet, B^\bullet)$ is defined by

$$\mathrm{Hom}^n(A^\bullet, B^\bullet) = \prod_{i} \mathrm{Hom}(A^i, B^{n+i})$$

with the differential $d^n$ is defined as follows. Given a collection of maps $\phi^i \in \mathrm{Hom}(A^i, B^{n+i})$ defining an element $(\phi^i)_i \in \mathrm{Hom}^n(A^\bullet, B^\bullet)$, the differential $d^n$ maps it to $(d_B^{n+i} \phi^i - (-1)^n \phi^{i+1} d_A^i)_i \in \mathrm{Hom}^{n+1}(A^\bullet, B^\bullet)$. There are probably a variety of ways to motivate this definition. Here are some.

(I) Think about the double complex $C^{p,q} = \mathrm{Hom}(A^{-p}, B^q)$. The vertical differential

$$\mathrm{Hom}(A^{-p}, B^q) \to \mathrm{Hom}(A^{-p}, B^{q+1})$$

is postcomposition with the differential $d_B^q : B^q \to B^{q+1}$, and the horizontal differential

$$\mathrm{Hom}(A^{-p}, B^q) \to \mathrm{Hom}(A^{-(p+1)}, B^q)$$

is precomposition with the differential $d_A^{-p} : A^{-p} \to A^{-(p+1)}$. I think this is a fairly natural double complex to write down. But the squares in this double complex commute, and usually we want squares in double complexes to anti-commute instead, so we should adjust our differentials by introducing a sign. Replace the horizontal differential $C^{p,q} \to C^{p+1, q}$ by $(-1)^{p+q+1}$ times precomposition with $d_A^{-p}$. If you now chase through the definition, I think you'll find that the associated total product complex of this double complex $C^{\bullet, \bullet}$ is precisely the complex $\mathrm{Hom}^\bullet(A^\bullet, B^\bullet)$ as we defined it above.

(II) Maybe the reason in (I) is not quite satisfying, since the sign we introduced seems a little arbitrary, and even ignoring that, at the very end we took the total product complex, and instead we could have taken the total coproduct complex. Here's one reason to motivate why the total product complex is the "right" one to take: given complexes $A^\bullet$ and $B^\bullet$, we have

$$Z^0 \mathrm{Hom}^\bullet(A^\bullet, B^\bullet) = \mathrm{Hom}_{\mathbf{C}}(A^\bullet, B^\bullet),$$

where I'm writing $\mathbf{C}$ for the category of complexes. Proving this is an easy exercise in understanding the definition of the differential. This would no longer be true if we had used the total coproduct complex instead, since morphisms of complexes aren't required to be 0 in all but finitely many degrees. What this tells us is that the category of complexes is enriched over the category of complexes of abelian groups, and that applying the functor $Z^0$ to these enriched hom's recovers the hom's of the usual preadditive category of complexes.

A related point is that a collection of maps $\phi^i \in \mathrm{Hom}(A^i, B^{0+i})$ defines an element of $B^0 \mathrm{Hom}^\bullet(A^\bullet, B^\bullet)$ if and only if it defines a null-homotopic morphism of complexes. This means that

$$H^0 \mathrm{Hom}^\bullet(A^\bullet, B^\bullet) = \mathrm{Hom}_{\mathbf{K}}(A^\bullet, B^\bullet),$$

where $\mathbf{K}$ is the "homotopy category of complexes," by which we mean the category whose objects are complexes whose morphisms are homotopy classes of morphisms of complexes.

(III) Here's the final motivating reason for the definition of the hom complex I can give, which also motivates the definition of the tensor product of complexes. Suppose we're working over a commutative ring $R$, so that the hom complex is also a complex of $R$-modules. The definitions of the hom complex and the tensor product of complexes give rise to an adjunction which makes the category of complexes of $R$-modules a closed monoidal category. More precisely, the statement is that for any triple of complexes $A^\bullet$, $B^\bullet$ and $C^\bullet$, there is an isomorphism of $R$-modules

$$\mathrm{Hom}_{\mathbf{Ch}}(A^\bullet \otimes B^\bullet, C^\bullet) \stackrel{\sim}{\rightarrow} \mathrm{Hom}_{\mathbf{Ch}}(A^\bullet, \mathrm{Hom}^\bullet(B^\bullet, C^\bullet))$$

which is natural in all three arguments. In fact, one can even show more generally that there is an isomorphism of complexes of $R$-modules

$$\mathrm{Hom}^{\bullet}(A^\bullet \otimes B^\bullet, C^\bullet) \stackrel{\sim}{\rightarrow} \mathrm{Hom}^\bullet(A^\bullet, \mathrm{Hom}^\bullet(B^\bullet, C^\bullet))$$

which is again natural in all three arguments (this is more general due to our observation in (II) above). The proof of this fact is annoying for notational reasons and is probably not worth doing in detail here.