The Monoidal Structure of Ch(R): Total Complexes, Tensor Ideals, and Internal Homs

The Tensor Product of Chain Complexes as a Total ComplexThe Double Complex Behind $C\otimes_R D$Totalizing the Double ComplexWhy This Sign WorksDirect Verification of $d^2=0$A Small Example: Tensoring Two Two-Term ComplexesThe Unit ComplexAssociativityThe Koszul SymmetryChecking That the Symmetry Is a Chain MapTensor Product of Chain MapsA Note on Homogeneous MapsCompatibility with Chain HomotopyThe Final PictureNull-Homotopic Maps Form a Tensor IdealAdditivityClosure Under CompositionClosure Under Tensoring on the RightClosure Under Tensoring on the LeftThe Tensor Ideal StatementThe Internal Hom ComplexWhy ${\delta }^2=0$Chain Maps and Homotopies Inside the Hom ComplexThe Tensor-Hom AdjunctionCompatibility with DifferentialsOrdinary Adjunction and Homotopy AdjunctionSummary

 

The Tensor Product of Chain Complexes as a Total Complex

A good way to understand the monoidal structure on $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ is not to start with abstract coherence diagrams, but with a much more concrete picture:

$$C{\otimes }_{R}D\text{is the total complex of the double complex}{E}_{p,q}=C_p{\otimes }_R{D}_q.$$

This point of view explains both parts of the definition:

$$(C{\otimes }_{R}D{)}_n=\underset{p+q=n}{⨁}{C}_p{\otimes }_R{D}_q,$$

and

$$d(c\otimes x)=d_{C}c\otimes x+(-1{)}^{|c|}c\otimes d_{D}x.$$

The first formula says that total degree is obtained by adding bidegrees. The second formula is the sign convention that makes the total differential square to zero.

Throughout, $R$$R$ is a commutative ring, and $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ means the category of homologically graded chain complexes of $R$$R$-modules:

$$\cdots ⟶C_{n+1}\stackrel{d_C}{\to }{C}_n\stackrel{d_C}{\to }{C}_{n-1}⟶\cdots$$

So the differential has degree $-1$$-1$.

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The Double Complex Behind $C{\otimes }_{R}D$$C\otimes_R D$

Given two chain complexes $C$$C$ and $D$$D$, form the bigraded object

$$E_{p,q}=C_p{\otimes }_R{D}_q.$$

There are two evident differentials. The horizontal one is induced by the differential of $C$$C$:

$$d^h=d_C\otimes 1:E_{p,q}⟶E_{p-1,q}.$$

The vertical one is induced by the differential of $D$$D$:

$$d^v=1\otimes d_D:E_{p,q}⟶E_{p,q-1}.$$

On the nose, these two maps commute:

$$d^h{d}^v=d^v{d}^h.$$

Indeed,

$$(d_C\otimes 1)(1\otimes d_D)=d_C\otimes d_D=(1\otimes d_D)(d_C\otimes 1).$$

But for a total complex, commuting differentials are not what we want. If one simply tried

$$d_{\mathrm{tot}}=d^h+d^v,$$

then

$$d_{\mathrm{tot}}^2=(d^h{)}^2+d^h{d}^v+d^v{d}^h+(d^v{)}^2.$$

The first and last terms vanish, but the middle terms add:

$$d_{\mathrm{tot}}^2=d^h{d}^v+d^v{d}^h.$$

Since $d^h{d}^v=d^v{d}^h$$d^h d^v=d^v d^h$, this is generally not zero.

So the total differential cannot be the naive sum. A sign has to be inserted.

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Totalizing the Double Complex

The tensor product complex $C{\otimes }_{R}D$$C\otimes_R D$ is defined by taking direct sums along diagonals:

$$(C{\otimes }_{R}D{)}_n=\underset{p+q=n}{⨁}{E}_{p,q}=\underset{p+q=n}{⨁}{C}_p{\otimes }_R{D}_q.$$

Thus, if $c\in C_p$$c\in C_p$ and $x\in D_q$$x\in D_q$, then

$$c\otimes x\in (C{\otimes }_{R}D{)}_{p+q}.$$

The total differential is

$$d_{\mathrm{tot}}=d^h+(-1{)}^p{d}^v\text{on }{E}_{p,q}.$$

Equivalently,

$$d(c\otimes x)=d_{C}c\otimes x+(-1{)}^{p}c\otimes d_{D}x,$$

where $p=|c|$$p=|c|$.

This is the usual formula for the tensor product of chain complexes.

The sign depends on the degree of the left factor. Conceptually, the vertical differential has to pass across the horizontal degree $p$$p$, and the price is the sign $(-1{)}^p$$(-1)^p$.

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Why This Sign Works

Let

$${\tilde{d}}^v=(-1{)}^p{d}^v\text{on }{E}_{p,q}.$$

Then

$$d_{\mathrm{tot}}=d^h+{\tilde{d}}^v.$$

We want the two pieces to anticommute:

$$d^h{\tilde{d}}^v+{\tilde{d}}^v{d}^h=0.$$

Compute on $E_{p,q}$$E_{p,q}$. First,

$$d^h{\tilde{d}}^v=d^h((-1{)}^p{d}^v)=(-1{)}^p{d}^h{d}^v.$$

On the other hand, after applying $d^h$$d^h$, the first index drops from $p$$p$ to $p-1$$p-1$. Hence

$${\tilde{d}}^v{d}^h=(-1{)}^{p-1}{d}^v{d}^h.$$

Since $d^h{d}^v=d^v{d}^h$$d^h d^v=d^v d^h$, we get

$$d^h{\tilde{d}}^v+{\tilde{d}}^v{d}^h=((-1{)}^p+(-1{)}^{p-1})d^h{d}^v=0.$$

Therefore

$$d_{\mathrm{tot}}^2=0.$$

That is the whole reason for the sign.

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Direct Verification of $d^2=0$$d^2=0$

Let $c\in C_p$$c\in C_p$ and $x\in D_q$$x\in D_q$. Then

$$d(c\otimes x)=d_{C}c\otimes x+(-1{)}^{p}c\otimes d_{D}x.$$

Apply $d$$d$ again. The first term gives

$$d(d_{C}c\otimes x)=d_C^{2}c\otimes x+(-1{)}^{p-1}{d}_{C}c\otimes d_{D}x.$$

The second term gives

$$d((-1{)}^{p}c\otimes d_{D}x)=(-1{)}^p{d}_{C}c\otimes d_{D}x+(-1{)}^p(-1{)}^{p}c\otimes d_D^{2}x.$$

Thus

$$d^2(c\otimes x)=d_C^{2}c\otimes x+((-1{)}^{p-1}+(-1{)}^p)d_{C}c\otimes d_{D}x+c\otimes d_D^{2}x.$$

Since $d_C^2=0$$d_C^2=0$ and $d_D^2=0$$d_D^2=0$, and since

$$(-1{)}^{p-1}+(-1{)}^p=0,$$

we get

$$d^2(c\otimes x)=0.$$

Without the sign $(-1{)}^p$$(-1)^p$, the two middle terms would add instead of canceling.

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A Small Example: Tensoring Two Two-Term Complexes

Let

$$P=(0\to R\stackrel{a}{\to }R\to 0),$$

and

$$Q=(0\to R\stackrel{b}{\to }R\to 0),$$

with nonzero terms in degrees $1$$1$ and $0$$0$.

Then $P{\otimes }_{R}Q$$P\otimes_R Q$ has nonzero terms in degrees $2,1,0$$2,1,0$.

In degree $2$$2$,

$$(P\otimes Q{)}_2=P_1\otimes Q_1\cong R.$$

In degree $1$$1$,

$$(P\otimes Q{)}_1=(P_0\otimes Q_1)\oplus (P_1\otimes Q_0)\cong R\oplus R.$$

In degree $0$$0$,

$$(P\otimes Q{)}_0=P_0\otimes Q_0\cong R.$$

So $P{\otimes }_{R}Q$$P\otimes_R Q$ has the form

$$0\to R\stackrel{d_2}{\to }R\oplus R\stackrel{d_1}{\to }R\to 0.$$

With the above ordering of the middle term, the first differential is

$$d_2(1)=(a,-b).$$

Indeed,

$$d(1\otimes 1)=d_P(1)\otimes 1-1\otimes d_Q(1).$$

The second differential is

$$d_1(x,y)=bx+ay.$$

Therefore

$$d_1{d}_2(1)=d_1(a,-b)=ba-ab=0.$$

This is the smallest place where the sign becomes visible. The minus sign is not cosmetic; it is exactly what makes the composition vanish.

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The Unit Complex

The tensor unit in $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ is $R[0]$$R[0]$, the complex with $R$$R$ in degree $0$$0$ and zero elsewhere:

$$\cdots \to 0\to R\to 0\to \cdots$$

Then

$$(C{\otimes }_{R}R[0]{)}_n=\underset{p+q=n}{⨁}{C}_p{\otimes }_{R}R[0{]}_q.$$

Only the summand with $q=0$$q=0$ survives, so

$$(C{\otimes }_{R}R[0]{)}_n\cong C_n{\otimes }_{R}R\cong C_n.$$

Similarly,

$$(R[0]{\otimes }_{R}C{)}_n\cong R{\otimes }_R{C}_n\cong C_n.$$

Thus

$$C{\otimes }_{R}R[0]\cong C,R[0]{\otimes }_{R}C\cong C.$$

So $R[0]$$R[0]$ is the monoidal unit.

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Associativity

There is a natural isomorphism

$$(C{\otimes }_{R}D){\otimes }_{R}E\cong C{\otimes }_R(D{\otimes }_{R}E).$$

On the underlying graded modules, both sides have degree $n$$n$ part

$$\underset{p+q+r=n}{⨁}{C}_p{\otimes }_R{D}_q{\otimes }_R{E}_r.$$

The differential on a pure tensor $c\otimes x\otimes y$$c\otimes x\otimes y$, with $c\in C_p$$c\in C_p$ and $x\in D_q$$x\in D_q$, is

$$d(c\otimes x\otimes y)=d_{C}c\otimes x\otimes y+(-1{)}^{p}c\otimes d_{D}x\otimes y+(-1{)}^{p+q}c\otimes x\otimes d_{E}y.$$

This is just the same total-complex rule, now for a triple complex.

So the associativity isomorphism is not merely an isomorphism of graded modules; it is an isomorphism of chain complexes.

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The Koszul Symmetry

Since $R$$R$ is commutative, one expects a symmetry

$$C{\otimes }_{R}D\cong D{\otimes }_{R}C.$$

But in chain complexes the symmetry is not the naive map $c\otimes x\mapsto x\otimes c$$c\otimes x\mapsto x\otimes c$. The correct formula is

$${\tau }_{C,D}(c\otimes x)=(-1{)}^{|c||x|}x\otimes c.$$

This is the Koszul sign rule.

It says that when an element of degree $p$$p$ crosses an element of degree $q$$q$, the sign is

$$(-1{)}^{pq}.$$

Odd-degree elements anticommute with odd-degree elements. Everything else commutes.

This sign is forced by the condition that ${\tau }_{C,D}$$\tau_{C,D}$ be a chain map.

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Checking That the Symmetry Is a Chain Map

Let $c\in C_p$$c\in C_p$ and $x\in D_q$$x\in D_q$.

First compute $d\tau (c\otimes x)$$d\tau(c\otimes x)$:

$$d\tau (c\otimes x)=d((-1{)}^{pq}x\otimes c).$$

Using the differential on $D\otimes C$$D\otimes C$, this is

$$(-1{)}^{pq}(d_{D}x\otimes c+(-1{)}^{q}x\otimes d_{C}c).$$

So

$$d\tau (c\otimes x)=(-1{)}^{pq}{d}_{D}x\otimes c+(-1{)}^{pq+q}x\otimes d_{C}c.$$

Now compute $\tau d(c\otimes x)$$\tau d(c\otimes x)$:

$$\tau d(c\otimes x)=\tau (d_{C}c\otimes x+(-1{)}^{p}c\otimes d_{D}x).$$

For the first term, $d_{C}c$$d_Cc$ has degree $p-1$$p-1$, so

$$\tau (d_{C}c\otimes x)=(-1{)}^{(p-1)q}x\otimes d_{C}c.$$

For the second term, $d_{D}x$$d_Dx$ has degree $q-1$$q-1$, so

$$\tau ((-1{)}^{p}c\otimes d_{D}x)=(-1{)}^p(-1{)}^{p(q-1)}{d}_{D}x\otimes c.$$

Hence

$$\tau d(c\otimes x)=(-1{)}^{(p-1)q}x\otimes d_{C}c+(-1{)}^{p+p(q-1)}{d}_{D}x\otimes c.$$

Modulo $2$$2$,

$$(p-1)q\equiv pq+q,$$

and

$$p+p(q-1)\equiv pq.$$

Therefore

$$\tau d(c\otimes x)=(-1{)}^{pq+q}x\otimes d_{C}c+(-1{)}^{pq}{d}_{D}x\otimes c.$$

This agrees with $d\tau (c\otimes x)$$d\tau(c\otimes x)$.

So $\tau$$\tau$ is a chain map.

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Tensor Product of Chain Maps

If

$$f:C\to C^{\prime },g:D\to D^{\prime }$$

are chain maps, then

$$f\otimes g:C{\otimes }_{R}D\to C^{\prime }{\otimes }_R{D}^{\prime }$$

is defined on pure tensors by

$$(f\otimes g)(c\otimes x)=f(c)\otimes g(x).$$

Because $f$$f$ and $g$$g$ have degree $0$$0$, there is no extra sign here.

The verification that $f\otimes g$$f\otimes g$ is a chain map is straightforward:

$$d(f\otimes g)(c\otimes x)=d(f(c)\otimes g(x)).$$

Since $f(c)$$f(c)$ has the same degree as $c$$c$,

$$d(f(c)\otimes g(x))=d_{C^{\prime }}f(c)\otimes g(x)+(-1{)}^{|c|}f(c)\otimes d_{D^{\prime }}g(x).$$

Using

$$d_{C^{\prime }}f=f{d}_C,d_{D^{\prime }}g=g{d}_D,$$

we get

$$d(f\otimes g)(c\otimes x)=f(d_{C}c)\otimes g(x)+(-1{)}^{|c|}f(c)\otimes g(d_{D}x).$$

This is exactly

$$(f\otimes g)d(c\otimes x).$$

Hence

$$d(f\otimes g)=(f\otimes g)d.$$

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A Note on Homogeneous Maps

In the ordinary category $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$, morphisms are degree $0$$0$ chain maps. But once one works with Hom complexes or dg categories, one also considers homogeneous maps of arbitrary degree.

If $f$$f$ has degree $|f|$$|f|$ and $g$$g$ has degree $|g|$$|g|$, the tensor product of homogeneous maps is defined by

$$(f\otimes g)(c\otimes x)=(-1{)}^{|g||c|}f(c)\otimes g(x).$$

The sign appears because $g$$g$ has to pass across the element $c$$c$ before acting on $x$$x$.

With this convention, one may write the differential on $C\otimes D$$C\otimes D$ formally as

$$d_{C\otimes D}=d_C\otimes 1+1\otimes d_D.$$

The sign in the second term is already built into the tensor product of homogeneous maps.

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Compatibility with Chain Homotopy

The tensor product also behaves correctly with chain homotopy.

Suppose

$$f:C\to D$$

is null-homotopic. Thus there is a degree $+1$$+1$ map

$$s:C\to D$$

such that

$$f=d_{D}s+s{d}_C.$$

Then for any complex $Z$$Z$,

$$f\otimes 1_Z\simeq 0,$$

and

$$1_Z\otimes f\simeq 0.$$

For the first statement, define

$$H(c\otimes z)=s(c)\otimes z.$$

A direct computation gives

$$dH+Hd=f\otimes 1_Z.$$

For the second statement, the homotopy needs a sign:

$$K(z\otimes c)=(-1{)}^{|z|}z\otimes s(c).$$

Then

$$dK+Kd=1_Z\otimes f.$$

Thus null-homotopic maps are closed under tensoring on either side.

Consequently, null-homotopic maps form a tensor ideal in $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$.

This is why the tensor product descends to the homotopy category

$$K(R)=\mathrm{Ch}(R)/\{\text{null-homotopic maps}\}.$$

The formula is simply

$$[f]\otimes [g]=[f\otimes g].$$

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The Final Picture

The monoidal structure on $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ can be read from one construction:

$$E_{p,q}=C_p{\otimes }_R{D}_q.$$

Then

$$C{\otimes }_{R}D=\mathrm{Tot}(E),$$

with

$$(C{\otimes }_{R}D{)}_n=\underset{p+q=n}{⨁}{E}_{p,q}.$$

The total differential is

$$d_{\mathrm{tot}}=d^h+(-1{)}^p{d}^v.$$

In element notation,

$$d(c\otimes x)=d_{C}c\otimes x+(-1{)}^{|c|}c\otimes d_{D}x.$$

From this total-complex viewpoint, the usual features of $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ become natural:

$$R[0]\text{is the tensor unit,}$$

$$(C\otimes D)\otimes E\cong C\otimes (D\otimes E),$$

and, when $R$$R$ is commutative,

$$C\otimes D\cong D\otimes C$$

via

$$c\otimes x\mapsto (-1{)}^{|c||x|}x\otimes c.$$

So the slogan is:

$$\text{The tensor product of chain complexes is the total complex of the tensor-product double complex.}$$

The rest is bookkeeping: the bookkeeping of degrees, and the bookkeeping of signs.

Null-Homotopic Maps Form a Tensor Ideal

The tensor product of chain complexes is not merely compatible with chain maps. It is also compatible with chain homotopies.

The statement is:

$$\boxed{{\displaystyle \text{null-homotopic maps form a tensor ideal in }\mathrm{Ch}(R).}}$$

Let

$$\mathcal{N}(C,D)=\{f:C\to D\mid f\simeq 0\}.$$

Thus $f\in \mathcal{N}(C,D)$$f\in \mathcal N(C,D)$ means that there exists a degree $+1$$+1$ map

$$s:C\to D$$

such that

$$f=d_{D}s+s{d}_C.$$

In components, this says

$$f_n=d_{n+1}^D{s}_n+s_{n-1}{d}_n^C.$$

We shall check that $\mathcal{N}$$\mathcal N$ is closed under addition, under composition from both sides, and under tensoring from both sides.

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Additivity

If

$$f=d_{D}s+s{d}_C$$

and

$$g=d_{D}t+t{d}_C,$$

then

$$f+g=d_D(s+t)+(s+t)d_C.$$

Also,

$$-f=d_D(-s)+(-s)d_C.$$

So each $\mathcal{N}(C,D)$$\mathcal N(C,D)$ is a subgroup of ${\mathrm{Hom}}_{\mathrm{Ch}(R)}(C,D)$$\mathrm{Hom}_{\mathrm{Ch}(R)}(C,D)$.

---

Closure Under Composition

Suppose

$$f:C\to D$$

is null-homotopic, say

$$f=d_{D}s+s{d}_C.$$

Let

$$u:B\to C,v:D\to E$$

be arbitrary chain maps. Then

$$vfu=v(d_{D}s+s{d}_C)u.$$

Since $u$$u$ and $v$$v$ commute with differentials,

$$v{d}_D=d_{E}v,d_{C}u=u{d}_B.$$

Therefore

$$vfu=d_E(vsu)+(vsu)d_B.$$

So

$$vfu\simeq 0.$$

Thus null-homotopic maps form a two-sided ideal in the ordinary categorical sense.

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Closure Under Tensoring on the Right

Now let $Z$$Z$ be another chain complex. We prove that

$$f\otimes 1_Z:C\otimes Z\to D\otimes Z$$

is null-homotopic.

Define a degree $+1$$+1$ map

$$H:C\otimes Z\to D\otimes Z$$

by

$$H(c\otimes z)=s(c)\otimes z.$$

Let $c\in C_p$$c\in C_p$ and $z\in Z_q$$z\in Z_q$. Then

$$dH(c\otimes z)=d_{D}s(c)\otimes z+(-1{)}^{p+1}s(c)\otimes d_{Z}z.$$

On the other hand,

$$Hd(c\otimes z)=H(d_{C}c\otimes z+(-1{)}^{p}c\otimes d_{Z}z),$$

so

$$Hd(c\otimes z)=s{d}_C(c)\otimes z+(-1{)}^{p}s(c)\otimes d_{Z}z.$$

Adding the two expressions gives

$$(dH+Hd)(c\otimes z)=(d_{D}s+s{d}_C)(c)\otimes z.$$

The two terms involving $d_{Z}z$$d_Zz$ cancel:

$$(-1{)}^{p+1}s(c)\otimes d_{Z}z+(-1{)}^{p}s(c)\otimes d_{Z}z=0.$$

Hence

$$dH+Hd=f\otimes 1_Z.$$

Therefore

$$f\otimes 1_Z\simeq 0.$$

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Closure Under Tensoring on the Left

The other side is similar, but a sign is needed.

We want to prove that

$$1_Z\otimes f:Z\otimes C\to Z\otimes D$$

is null-homotopic.

Define

$$K:Z\otimes C\to Z\otimes D$$

by

$$K(z\otimes c)=(-1{)}^{|z|}z\otimes s(c).$$

Let $z\in Z_q$$z\in Z_q$ and $c\in C_p$$c\in C_p$. Then

$$dK(z\otimes c)=(-1{)}^q{d}_{Z}z\otimes s(c)+z\otimes d_{D}s(c).$$

Also,

$$d(z\otimes c)=d_{Z}z\otimes c+(-1{)}^{q}z\otimes d_{C}c.$$

Therefore

$$Kd(z\otimes c)=K(d_{Z}z\otimes c)+(-1{)}^{q}K(z\otimes d_{C}c).$$

Since $d_{Z}z$$d_Zz$ has degree $q-1$$q-1$,

$$K(d_{Z}z\otimes c)=(-1{)}^{q-1}{d}_{Z}z\otimes s(c).$$

And

$$(-1{)}^{q}K(z\otimes d_{C}c)=(-1{)}^q(-1{)}^{q}z\otimes s{d}_C(c)=z\otimes s{d}_C(c).$$

Thus

$$Kd(z\otimes c)=(-1{)}^{q-1}{d}_{Z}z\otimes s(c)+z\otimes s{d}_C(c).$$

Adding gives

$$(dK+Kd)(z\otimes c)=z\otimes (d_{D}s+s{d}_C)(c).$$

Again, the terms involving $d_{Z}z$$d_Zz$ cancel:

$$(-1{)}^q+(-1{)}^{q-1}=0.$$

So

$$dK+Kd=1_Z\otimes f.$$

Hence

$$1_Z\otimes f\simeq 0.$$

The sign

$$(-1{)}^{|z|}$$

in the definition of $K$$K$ is forced. Without it, the $d_Z$$d_Z$-terms would not cancel.

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The Tensor Ideal Statement

Combining the previous observations, null-homotopic maps satisfy:

$$f\simeq 0⟹vfu\simeq 0$$

for all composable chain maps $u,v$$u,v$, and

$$f\simeq 0⟹f\otimes 1_Z\simeq 0,1_Z\otimes f\simeq 0.$$

More generally, if

$$g:A\to B$$

is any chain map, then

$$f\otimes g=(1_D\otimes g)(f\otimes 1_A)$$

is null-homotopic, and

$$g\otimes f=(g\otimes 1_D)(1_A\otimes f)$$

is null-homotopic.

Thus

$$\boxed{{\displaystyle \mathcal{N}\text{ is a tensor ideal of morphisms in }\mathrm{Ch}(R).}}$$

Consequently, the tensor product descends to the homotopy category

$$K(R)=\mathrm{Ch}(R)/\mathcal{N}.$$

Indeed, if

$$f\simeq f^{\prime },g\simeq g^{\prime },$$

then

$$f\otimes g-f^{\prime }\otimes g^{\prime }=(f-f^{\prime })\otimes g+f^{\prime }\otimes (g-g^{\prime }).$$

Both terms on the right are null-homotopic. Hence

$$f\otimes g\simeq f^{\prime }\otimes g^{\prime }.$$

So the formula

$$[f]\otimes [g]=[f\otimes g]$$

is well-defined in $K(R)$$K(R)$.

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The Internal Hom Complex

The tensor product on $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ has a right adjoint. This is the internal Hom complex.

For chain complexes $C$$C$ and $D$$D$, define

$$\underset{―}{\mathrm{Hom}}(C,D{)}_n=\prod _{p\in \mathbb{Z}}{\mathrm{Hom}}_R(C_p,D_{p+n}).$$

Thus an element

$$\phi \in \underset{―}{\mathrm{Hom}}(C,D{)}_n$$

is a homogeneous map of degree $n$$n$:

$${\phi }_p:C_p\to D_{p+n}.$$

The differential is

$$\boxed{{\displaystyle \delta (\phi )=d_D\phi -(-1{)}^n\phi d_C.}}$$

In components,

$$(\delta \phi {)}_p=d_D\circ {\phi }_p-(-1{)}^n{\phi }_{p-1}\circ d_C.$$

Since $\phi$$\varphi$ has degree $n$$n$, the element $\delta (\phi )$$\delta(\varphi)$ has degree $n-1$$n-1$.

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Why ${\delta }^2=0$$\delta^2=0$

Let $|\phi |=n$$|\varphi|=n$. Then

$$\delta (\phi )=d_D\phi -(-1{)}^n\phi d_C.$$

Applying $\delta$$\delta$ again gives

$${\delta }^2(\phi )=d_D\delta (\phi )-(-1{)}^{n-1}\delta (\phi )d_C.$$

Substitute the formula for $\delta (\phi )$$\delta(\varphi)$:

$${\delta }^2(\phi )=d_D(d_D\phi -(-1{)}^n\phi d_C)-(-1{)}^{n-1}(d_D\phi -(-1{)}^n\phi d_C)d_C.$$

Expanding,

$${\delta }^2(\phi )=d_D^2\phi -(-1{)}^n{d}_D\phi d_C-(-1{)}^{n-1}{d}_D\phi d_C+(-1{)}^{2n-1}\phi d_C^2.$$

The first and last terms vanish because

$$d_D^2=0,d_C^2=0.$$

The middle terms cancel because

$$-(-1{)}^n-(-1{)}^{n-1}=0.$$

Therefore

$${\delta }^2(\phi )=0.$$

So $\underset{―}{\mathrm{Hom}}(C,D)$$\underline{\mathrm{Hom}}(C,D)$ is indeed a chain complex.

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Chain Maps and Homotopies Inside the Hom Complex

The definition of $\underset{―}{\mathrm{Hom}}(C,D)$$\underline{\mathrm{Hom}}(C,D)$ packages chain maps and chain homotopies into one complex.

First,

$$Z_0\underset{―}{\mathrm{Hom}}(C,D)$$

is the group of degree $0$$0$ cycles. A degree $0$$0$ element $\phi :C\to D$$\varphi:C\to D$ is a cycle precisely when

$$\delta (\phi )=0.$$

But

$$\delta (\phi )=d_D\phi -\phi d_C.$$

Thus

$$\delta (\phi )=0$$

if and only if

$$d_D\phi =\phi d_C.$$

So

$$Z_0\underset{―}{\mathrm{Hom}}(C,D)={\mathrm{Hom}}_{\mathrm{Ch}(R)}(C,D).$$

Next,

$$B_0\underset{―}{\mathrm{Hom}}(C,D)$$

consists of degree $0$$0$ boundaries. These come from degree $1$$1$ elements

$$s\in \underset{―}{\mathrm{Hom}}(C,D{)}_1.$$

For such an $s$$s$,

$$\delta (s)=d_{D}s-(-1{)}^{1}s{d}_C=d_{D}s+s{d}_C.$$

But this is exactly the form of a null-homotopic map.

Hence

$$B_0\underset{―}{\mathrm{Hom}}(C,D)=\{f:C\to D\mid f\simeq 0\}.$$

Therefore

$$\boxed{{\displaystyle H_0\underset{―}{\mathrm{Hom}}(C,D)\cong {\mathrm{Hom}}_{K(R)}(C,D).}}$$

This is the clean conceptual meaning of the homotopy category: its morphisms are the degree-zero homology classes of the mapping complex.

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The Tensor-Hom Adjunction

The internal Hom is right adjoint to tensor product.

There is a natural isomorphism of chain complexes

$$\boxed{{\displaystyle \underset{―}{\mathrm{Hom}}(C{\otimes }_{R}D,E)\cong \underset{―}{\mathrm{Hom}}(C,\underset{―}{\mathrm{Hom}}(D,E)).}}$$

This is the tensor-Hom adjunction in $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$.

At the level of homogeneous maps, the correspondence is the usual currying operation.

Let

$$F:C{\otimes }_{R}D\to E$$

be a homogeneous map of degree $n$$n$. Define

$$\hat{F}:C\to \underset{―}{\mathrm{Hom}}(D,E)$$

by

$$\hat{F}(c)(x)=F(c\otimes x).$$

If $c\in C_p$$c\in C_p$, then $\hat{F}(c)$$\widehat F(c)$ is a map

$$D_q\to E_{p+q+n}.$$

So $\hat{F}(c)$$\widehat F(c)$ has degree $p+n$$p+n$ as an element of $\underset{―}{\mathrm{Hom}}(D,E)$$\underline{\mathrm{Hom}}(D,E)$. Therefore $\hat{F}$$\widehat F$ itself has degree $n$$n$.

Conversely, given a homogeneous map

$$G:C\to \underset{―}{\mathrm{Hom}}(D,E)$$

of degree $n$$n$, define

$$\tilde{G}:C{\otimes }_{R}D\to E$$

by

$$\tilde{G}(c\otimes x)=G(c)(x).$$

These two constructions are inverse to each other.

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Compatibility with Differentials

The main point is that the currying isomorphism is not just an isomorphism of graded modules. It is an isomorphism of chain complexes.

Let $F$$F$ have degree $n$$n$, and let $c\in C_p$$c\in C_p$, $x\in D_q$$x\in D_q$.

First compute the differential after currying.

Since $\hat{F}(c)$$\widehat F(c)$ has degree $n+p$$n+p$ inside $\underset{―}{\mathrm{Hom}}(D,E)$$\underline{\mathrm{Hom}}(D,E)$,

$$(\delta \hat{F})(c)(x)=d_{E}F(c\otimes x)-(-1{)}^{n+p}F(c\otimes d_{D}x)-(-1{)}^{n}F(d_{C}c\otimes x).$$

On the other hand,

$$\widehat{\delta F}(c)(x)=(\delta F)(c\otimes x).$$

Since $F$$F$ has degree $n$$n$,

$$(\delta F)(c\otimes x)=d_{E}F(c\otimes x)-(-1{)}^{n}F(d_{C\otimes D}(c\otimes x)).$$

Using

$$d_{C\otimes D}(c\otimes x)=d_{C}c\otimes x+(-1{)}^{p}c\otimes d_{D}x,$$

we get

$$\widehat{\delta F}(c)(x)=d_{E}F(c\otimes x)-(-1{)}^{n}F(d_{C}c\otimes x)-(-1{)}^{n+p}F(c\otimes d_{D}x).$$

This agrees with $(\delta \hat{F})(c)(x)$$(\delta\widehat F)(c)(x)$.

Therefore currying commutes with differentials:

$$\widehat{\delta F}=\delta \hat{F}.$$

Thus

$$\underset{―}{\mathrm{Hom}}(C{\otimes }_{R}D,E)\cong \underset{―}{\mathrm{Hom}}(C,\underset{―}{\mathrm{Hom}}(D,E))$$

as chain complexes.

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Ordinary Adjunction and Homotopy Adjunction

Taking degree-zero cycles gives the ordinary adjunction in $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$:

$${\mathrm{Hom}}_{\mathrm{Ch}(R)}(C{\otimes }_{R}D,E)\cong {\mathrm{Hom}}_{\mathrm{Ch}(R)}(C,\underset{―}{\mathrm{Hom}}(D,E)).$$

Taking degree-zero homology gives the corresponding adjunction in the homotopy category:

$${\mathrm{Hom}}_{K(R)}(C{\otimes }_{R}D,E)\cong {\mathrm{Hom}}_{K(R)}(C,\underset{―}{\mathrm{Hom}}(D,E)).$$

So the closed monoidal structure is already visible before passing to derived categories.

In the derived category $D(R)$$D(R)$, one has to replace these by their derived versions:

$$-{\otimes }_R^{\mathbb{L}}-$$

and

$$\mathbb{R}\underset{―}{\mathrm{Hom}}(-,-).$$

But at the level of $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ and $K(R)$$K(R)$, the basic tensor-Hom adjunction is the one above.

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Summary

The monoidal structure on $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ has two companion facts.

First, null-homotopic maps form a tensor ideal:

$$f\simeq 0⟹f\otimes g\simeq 0,g\otimes f\simeq 0.$$

This is why the tensor product descends from $\mathrm{Ch}(R)$$\mathrm{Ch}(R)$ to $K(R)$$K(R)$.

Second, the tensor product has an internal right adjoint:

$$\underset{―}{\mathrm{Hom}}(C{\otimes }_{R}D,E)\cong \underset{―}{\mathrm{Hom}}(C,\underset{―}{\mathrm{Hom}}(D,E)).$$

The Hom complex also explains chain homotopy:

$$Z_0\underset{―}{\mathrm{Hom}}(C,D)={\mathrm{Hom}}_{\mathrm{Ch}(R)}(C,D),$$

while

$$B_0\underset{―}{\mathrm{Hom}}(C,D)=\{\text{null-homotopic maps }C\to D\}.$$

Hence

$$H_0\underset{―}{\mathrm{Hom}}(C,D)\cong {\mathrm{Hom}}_{K(R)}(C,D).$$

So the passage

$$\mathrm{Ch}(R)\to K(R)$$

is already encoded inside the internal Hom complex.