The Architecture Behind the Hom Differential II: Tensor-Hom and DG Enrichment

The Architecture Behind the Hom Differential II: Tensor-Hom and DG EnrichmentThe Tensor-Hom Adjunction in Chain ComplexesWhy Enrichment AppearsWhat a $\operatorname{Ch}(K)$-Enriched Category IsThe Hom Complex as the Enriched Hom ObjectDefining CompositionWhy the Leibniz Rule Means That Composition Is a Chain MapVerifying the Leibniz RuleThe Unit MapAssociativity and UnitalityTaking $Z_0$ Recovers Ordinary CompositionThe Homotopy Category as $H_0$Endomorphisms Form a DG AlgebraSummary

 

The Architecture Behind the Hom Differential II: Tensor-Hom and DG Enrichment

In the first part, we saw that tensor products and Hom complexes arise from the same general mechanism:

$$\text{bifunctors}⟶\text{signed double complexes}⟶\text{total complexes}.$$

The tensor product of chain complexes is direct-sum totalization. The Hom complex is product totalization. The Hom complex also contains chain maps and chain homotopies:

$$Z_0\underset{―}{\mathrm{Hom}}(C,D)=\text{chain maps},$$

$$H_0\underset{―}{\mathrm{Hom}}(C,D)=\text{chain maps modulo chain homotopy}.$$

Now we explain the next structural point.

The tensor product and Hom complex fit into the tensor-Hom adjunction in $\mathrm{Ch}(R)$$\operatorname{Ch}(R)$. Then the Hom complex upgrades the category of chain complexes into a dg category.

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The Tensor-Hom Adjunction in Chain Complexes

Let $R$$R$ be a commutative ring.

For chain complexes of $R$$R$-modules, the tensor product is

$$X{\otimes }_{R}Y={\mathrm{Tot}}^{\oplus }(X_p{\otimes }_R{Y}_q).$$

Thus

$$(X{\otimes }_{R}Y{)}_n=\underset{p+q=n}{⨁}{X}_p{\otimes }_R{Y}_q.$$

The differential is

$$d(x\otimes y)=d_{X}x\otimes y+(-1{)}^{|x|}x\otimes d_{Y}y.$$

The Hom complex is

$${\underset{―}{\mathrm{Hom}}}_R(Y,Z{)}_n=\prod _j{\mathrm{Hom}}_R(Y_j,Z_{j+n}),$$

with differential

$$(df{)}_j=d_Z{f}_j-(-1{)}^{|f|}{f}_{j-1}{d}_Y.$$

These two constructions satisfy the closed monoidal adjunction

$${\underset{―}{\mathrm{Hom}}}_R(X{\otimes }_{R}Y,Z)\cong {\underset{―}{\mathrm{Hom}}}_R(X,{\underset{―}{\mathrm{Hom}}}_R(Y,Z)).$$

This is an isomorphism of chain complexes.

On degree $n$$n$ elements, the isomorphism is ordinary currying.

A degree $n$$n$ element on the left is a family of maps

$$\varphi =({\varphi }_{a,b}:X_a{\otimes }_R{Y}_b\to Z_{a+b+n}{)}_{a,b}.$$

It corresponds to a degree $n$$n$ element on the right, namely a family

$$\tilde{\varphi }=({\tilde{\varphi }}_a:X_a\to {\underset{―}{\mathrm{Hom}}}_R(Y,Z{)}_{a+n}{)}_a.$$

For

$$x\in X_a,$$

the element ${\tilde{\varphi }}_a(x)$$\widetilde{\phi}_a(x)$ is a degree $a+n$$a+n$ map from $Y$$Y$ to $Z$$Z$. Its $b$$b$-th component is

$$Y_b\to Z_{b+a+n},$$

given by

$$y\mapsto {\varphi }_{a,b}(x\otimes y).$$

Thus

$${\tilde{\varphi }}_a(x)(y)={\varphi }_{a,b}(x\otimes y).$$

The reason this is an isomorphism of chain complexes is that the tensor differential and the Hom differential are compatible.

The evaluation map

$$\mathrm{ev}:{\underset{―}{\mathrm{Hom}}}_R(Y,Z){\otimes }_{R}Y\to Z$$

is given without extra signs:

$$\mathrm{ev}(f\otimes y)=f(y).$$

Let $f$$f$ be homogeneous of degree $m$$m$. The tensor differential gives

$$d(f\otimes y)=df\otimes y+(-1{)}^{m}f\otimes d_{Y}y.$$

Applying evaluation gives

$$(df)(y)+(-1{)}^{m}f(d_{Y}y).$$

Using the Hom differential

$$df=d_{Z}f-(-1{)}^{m}f{d}_Y,$$

this becomes

$$d_Z(f(y))-(-1{)}^{m}f(d_{Y}y)+(-1{)}^{m}f(d_{Y}y).$$

The middle terms cancel, leaving

$$d_Z(f(y)).$$

Thus evaluation is a chain map.

This is the concrete reason the tensor-Hom adjunction lives in $\mathrm{Ch}(R)$$\operatorname{Ch}(R)$, not merely in graded $R$$R$-modules.

So $\mathrm{Ch}(R)$$\operatorname{Ch}(R)$ is a closed monoidal category:

$$-{\otimes }_{R}Y⊣{\underset{―}{\mathrm{Hom}}}_R(Y,-).$$

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Why Enrichment Appears

The Hom complex does more than package chain maps and chain homotopies. It upgrades the whole category of chain complexes into an enriched category.

The guiding idea is simple.

Ordinary categories have Hom sets:

$$\mathrm{Hom}(C,D).$$

A dg category has Hom complexes:

$$\mathrm{Map}(C,D)\in \mathrm{Ch}(K).$$

So in a dg category, morphisms are not merely elements of a set. They live inside chain complexes. Composition must therefore also live inside the world of chain complexes.

This is the point that needs to be made precise.

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What a $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$-Enriched Category Is

Recall that

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

is a monoidal category under the tensor product of chain complexes, with unit object

$$K[0].$$

A category enriched over $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$ consists of the following data.

First, a class of objects.

Second, for every pair of objects $A,B$$A,B$, a chain complex

$$\mathrm{Map}(A,B)\in \mathrm{Ch}(K).$$

Third, for every triple of objects $A,B,C$$A,B,C$, a composition morphism in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$:

$$\circ :\mathrm{Map}(B,C)\otimes \mathrm{Map}(A,B)\to \mathrm{Map}(A,C).$$

This is not just a bilinear map of graded modules. It must be a chain map.

Fourth, for every object $A$$A$, a unit morphism in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$:

$$K[0]\to \mathrm{Map}(A,A).$$

These composition and unit maps must satisfy associativity and unitality, now as identities of morphisms in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$.

A dg category is precisely a category enriched over $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$.

So to prove that

$$\mathrm{Ch}(\mathcal{C})$$

is a dg category, we must define:

$$\mathrm{Map}(C,D),$$

define composition as a chain map,

$$\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)\to \mathrm{Map}(C,E),$$

define the unit map,

$$K[0]\to \mathrm{Map}(C,C),$$

and check associativity and unit laws.

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The Hom Complex as the Enriched Hom Object

Let $\mathcal{C}$$\mathcal C$ be a $K$$K$-linear additive category.

For chain complexes

$$C,D\in \mathrm{Ch}(\mathcal{C}),$$

define

$$\mathrm{Map}(C,D)=\underset{―}{\mathrm{Hom}}(C,D).$$

Explicitly,

$$\mathrm{Map}(C,D{)}_n=\prod _i{\mathrm{Hom}}_{\mathcal{C}}(C_i,D_{i+n}).$$

A homogeneous element

$$f\in \mathrm{Map}(C,D{)}_n$$

is a family of morphisms

$$f=(f_i:C_i\to D_{i+n}{)}_i.$$

The differential is

$$(df{)}_i=d_D{f}_i-(-1{)}^n{f}_{i-1}{d}_C.$$

Thus the enriched Hom object from $C$$C$ to $D$$D$ is exactly the Hom complex.

Notice the first important consequence:

$$Z_0\mathrm{Map}(C,D)$$

is the set of ordinary chain maps $C\to D$$C\to D$.

Indeed, a degree-zero element is a family

$$f=(f_i:C_i\to D_i{)}_i,$$

and the condition $df=0$$df=0$ says

$$d_D{f}_i=f_{i-1}{d}_C,$$

which is precisely the chain map condition.

So the ordinary Hom set in $\mathrm{Ch}(\mathcal{C})$$\operatorname{Ch}(\mathcal C)$ is recovered as

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{C})}(C,D)=Z_0\mathrm{Map}(C,D).$$

This explains why degree-zero cycles, rather than all degree-zero elements, give the ordinary morphisms.

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Defining Composition

Now take three chain complexes

$$C,D,E\in \mathrm{Ch}(\mathcal{C}).$$

We want to define a composition morphism in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$:

$$\circ :\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)\to \mathrm{Map}(C,E).$$

Let

$$f=(f_i:C_i\to D_{i+n}{)}_i\in \mathrm{Map}(C,D{)}_n,$$

and

$$g=(g_j:D_j\to E_{j+m}{)}_j\in \mathrm{Map}(D,E{)}_m.$$

Thus $f$$f$ is a degree $n$$n$ family of morphisms, and $g$$g$ is a degree $m$$m$ family of morphisms.

To compose them, start with $C_i$$C_i$:

$$C_i\stackrel{f_i}{\to }{D}_{i+n}.$$

Now the relevant component of $g$$g$ is the one whose source is $D_{i+n}$$D_{i+n}$, namely

$$g_{i+n}:D_{i+n}\to E_{i+n+m}.$$

Therefore define

$$(g\circ f{)}_i=g_{i+n}\circ f_i.$$

Thus

$$g\circ f=(g_{i+n}\circ f_i:C_i\to E_{i+n+m}{)}_i.$$

So

$$g\circ f\in \mathrm{Map}(C,E{)}_{m+n}.$$

Composition adds degrees:

$$|g\circ f|=|g|+|f|.$$

This defines a graded bilinear map

$$\mathrm{Map}(D,E{)}_m\otimes \mathrm{Map}(C,D{)}_n\to \mathrm{Map}(C,E{)}_{m+n}.$$

Equivalently, it defines a morphism of graded $K$$K$-modules

$$\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)\to \mathrm{Map}(C,E).$$

But for enrichment over $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$, this is not enough. We must prove that this map is a chain map.

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Why the Leibniz Rule Means That Composition Is a Chain Map

Let

$$A=\mathrm{Map}(D,E),B=\mathrm{Map}(C,D),M=\mathrm{Map}(C,E).$$

The enriched composition is supposed to be a morphism in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$:

$$\mu :A\otimes B\to M.$$

Here

$$\mu (g\otimes f)=g\circ f.$$

So the source complex is

$$\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D),$$

and the target complex is

$$\mathrm{Map}(C,E).$$

A morphism in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$ is a chain map. Therefore $\mu$$\mu$ must satisfy

$$d_M\mu =\mu d_{A\otimes B}.$$

Equivalently, for every homogeneous simple tensor

$$g\otimes f\in A_m\otimes B_n,$$

we need

$$d_M(\mu (g\otimes f))=\mu (d_{A\otimes B}(g\otimes f)).$$

The left-hand side is

$$d_M(\mu (g\otimes f))=d_M(g\circ f)=d(g\circ f).$$

Now compute the right-hand side.

The tensor product differential on

$$A\otimes B=\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)$$

is

$$d_{A\otimes B}(g\otimes f)=dg\otimes f+(-1{)}^{m}g\otimes df,$$

where

$$m=|g|.$$

Applying $\mu$$\mu$ gives

$$\mu (d_{A\otimes B}(g\otimes f))=\mu (dg\otimes f)+(-1{)}^m\mu (g\otimes df).$$

By definition of $\mu$$\mu$,

$$\mu (dg\otimes f)=dg\circ f,$$

and

$$\mu (g\otimes df)=g\circ df.$$

Thus

$$\mu (d_{A\otimes B}(g\otimes f))=dg\circ f+(-1{)}^{m}g\circ df.$$

Therefore the chain map condition

$$d_M\mu =\mu d_{A\otimes B}$$

is exactly the formula

$$d(g\circ f)=dg\circ f+(-1{)}^{m}g\circ df.$$

This is the graded Leibniz rule for composition.

So the role of the Leibniz rule is not mysterious. It is simply the chain map condition for the composition map

$$\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)\to \mathrm{Map}(C,E)$$

after expanding the tensor product differential on the source.

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Verifying the Leibniz Rule

Now we verify

$$d(g\circ f)=dg\circ f+(-1{)}^{m}g\circ df.$$

Let

$$f=(f_i:C_i\to D_{i+n}{)}_i\in \mathrm{Map}(C,D{)}_n,$$

and

$$g=(g_j:D_j\to E_{j+m}{)}_j\in \mathrm{Map}(D,E{)}_m.$$

Then

$$g\circ f=(g_{i+n}{f}_i:C_i\to E_{i+n+m}{)}_i.$$

Since $g\circ f$$g\circ f$ has degree $m+n$$m+n$, its differential is

$$d(g\circ f{)}_i=d_E(g_{i+n}{f}_i)-(-1{)}^{m+n}(g\circ f{)}_{i-1}{d}_C.$$

Now

$$(g\circ f{)}_{i-1}=g_{i+n-1}{f}_{i-1}.$$

Thus

$$d(g\circ f{)}_i=d_E{g}_{i+n}{f}_i-(-1{)}^{m+n}{g}_{i+n-1}{f}_{i-1}{d}_C.$$

On the other hand,

$$(dg\circ f{)}_i=(dg{)}_{i+n}{f}_i.$$

Since $g$$g$ has degree $m$$m$,

$$(dg{)}_{i+n}=d_E{g}_{i+n}-(-1{)}^m{g}_{i+n-1}{d}_D.$$

Hence

$$(dg\circ f{)}_i=d_E{g}_{i+n}{f}_i-(-1{)}^m{g}_{i+n-1}{d}_D{f}_i.$$

Also,

$$(g\circ df{)}_i=g_{i+n-1}(df{)}_i.$$

Since $f$$f$ has degree $n$$n$,

$$(df{)}_i=d_D{f}_i-(-1{)}^n{f}_{i-1}{d}_C.$$

Therefore

$$(g\circ df{)}_i=g_{i+n-1}{d}_D{f}_i-(-1{)}^n{g}_{i+n-1}{f}_{i-1}{d}_C.$$

Multiplying by $(-1{)}^m$$(-1)^m$, we get

$$(-1{)}^m(g\circ df{)}_i=(-1{)}^m{g}_{i+n-1}{d}_D{f}_i-(-1{)}^{m+n}{g}_{i+n-1}{f}_{i-1}{d}_C.$$

Adding the two expressions gives

$$(dg\circ f{)}_i+(-1{)}^m(g\circ df{)}_i$$

$$=d_E{g}_{i+n}{f}_i-(-1{)}^m{g}_{i+n-1}{d}_D{f}_i+(-1{)}^m{g}_{i+n-1}{d}_D{f}_i-(-1{)}^{m+n}{g}_{i+n-1}{f}_{i-1}{d}_C.$$

The middle two terms cancel. Hence

$$(dg\circ f{)}_i+(-1{)}^m(g\circ df{)}_i=d_E{g}_{i+n}{f}_i-(-1{)}^{m+n}{g}_{i+n-1}{f}_{i-1}{d}_C.$$

This is exactly

$$d(g\circ f{)}_i.$$

Therefore

$$d(g\circ f)=dg\circ f+(-1{)}^{m}g\circ df.$$

So composition is a chain map in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$:

$$\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)\to \mathrm{Map}(C,E).$$

This is the enriched composition.

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The Unit Map

For each chain complex $C$$C$, the identity family

$${\mathrm{id}}_C=({\mathrm{id}}_{C_i}{)}_i$$

is a degree-zero element of

$$\mathrm{Map}(C,C).$$

It is a cycle because

$$(d{\mathrm{id}}_C{)}_i=d_C{\mathrm{id}}_{C_i}-{\mathrm{id}}_{C_{i-1}}{d}_C=d_C-d_C=0.$$

Therefore

$${\mathrm{id}}_C\in Z_0\mathrm{Map}(C,C).$$

Equivalently, it defines a chain map

$$K[0]\to \mathrm{Map}(C,C),$$

sending

$$1\mapsto {\mathrm{id}}_C.$$

This is the enriched identity.

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Associativity and Unitality

Associativity follows from ordinary associativity of composition in $\mathcal{C}$$\mathcal C$.

Indeed, if

$$f\in \mathrm{Map}(C,D),g\in \mathrm{Map}(D,E),h\in \mathrm{Map}(E,F),$$

then componentwise,

$$(h\circ (g\circ f){)}_i=h_{i+|f|+|g|}(g_{i+|f|}{f}_i),$$

while

$$((h\circ g)\circ f{)}_i=(h_{i+|f|+|g|}{g}_{i+|f|})f_i.$$

These are equal by associativity in $\mathcal{C}$$\mathcal C$.

Unitality follows similarly. For any homogeneous

$$f\in \mathrm{Map}(C,D),$$

we have

$$({\mathrm{id}}_D\circ f{)}_i={\mathrm{id}}_{D_{i+|f|}}{f}_i=f_i,$$

and

$$(f\circ {\mathrm{id}}_C{)}_i=f_i{\mathrm{id}}_{C_i}=f_i.$$

Thus the enriched composition is associative and unital.

Since the composition and unit maps are chain maps, these identities hold inside $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$.

Therefore

$$\mathrm{Ch}(\mathcal{C})$$

is enriched over

$$\mathrm{Ch}(K).$$

Equivalently,

$$\mathrm{Ch}(\mathcal{C})$$

is a dg category.

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Taking $Z_0$$Z_0$ Recovers Ordinary Composition

The ordinary category of chain complexes is obtained by taking degree-zero cycles of the enriched Hom complexes.

We have already seen that

$$Z_0\mathrm{Map}(C,D)={\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{C})}(C,D).$$

Now suppose

$$f\in Z_0\mathrm{Map}(C,D),g\in Z_0\mathrm{Map}(D,E).$$

Then

$$|f|=|g|=0,df=dg=0.$$

The enriched composition gives

$$(g\circ f{)}_i=g_i{f}_i.$$

This is exactly the ordinary componentwise composition of chain maps.

Moreover, since composition is a chain map,

$$d(g\circ f)=dg\circ f+g\circ df=0.$$

So

$$g\circ f\in Z_0\mathrm{Map}(C,E).$$

Thus applying $Z_0$$Z_0$ to the enriched composition recovers the usual composition law

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{C})}(D,E)\times {\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{C})}(C,D)\to {\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{C})}(C,E).$$

In this sense, ordinary composition is the degree-zero shadow of dg-enriched composition.

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The Homotopy Category as $H_0$$H_0$

Taking degree-zero homology gives the homotopy category.

A degree-one element

$$h\in \mathrm{Map}(C,D{)}_1$$

is a family

$$h=(h_i:C_i\to D_{i+1}{)}_i.$$

Its boundary is

$$(dh{)}_i=d_D{h}_i+h_{i-1}{d}_C.$$

This is precisely the formula for a null-homotopic chain map.

Therefore

$$H_0\mathrm{Map}(C,D)$$

is the module of chain maps modulo chain homotopy.

Hence

$${\mathrm{Hom}}_{\mathrm{K}(\mathcal{C})}(C,D)=H_0\mathrm{Map}(C,D).$$

So there are three layers:

$$\mathrm{Map}(C,D{)}_0=\text{graded maps},$$

$$Z_0\mathrm{Map}(C,D)=\text{chain maps},$$

$$H_0\mathrm{Map}(C,D)=\text{chain maps modulo chain homotopy}.$$

The dg-enriched category remembers the whole Hom complex. The ordinary category remembers only $Z_0$$Z_0$. The homotopy category remembers only $H_0$$H_0$.

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Endomorphisms Form a DG Algebra

In any enriched category, the endomorphism object of an object is automatically a monoid object.

Apply this to the $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$-enriched category

$$\mathrm{Ch}(\mathcal{C}).$$

For a chain complex $C$$C$, define

$$\underset{―}{\mathrm{End}}(C)=\mathrm{Map}(C,C)=\underset{―}{\mathrm{Hom}}(C,C).$$

Composition gives multiplication

$$\underset{―}{\mathrm{End}}(C)\otimes \underset{―}{\mathrm{End}}(C)\to \underset{―}{\mathrm{End}}(C),$$

and the identity gives the unit

$$K[0]\to \underset{―}{\mathrm{End}}(C).$$

Therefore

$$\underset{―}{\mathrm{End}}(C)$$

is a monoid object in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$.

But monoid objects in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$ are precisely differential graded $K$$K$-algebras.

Concretely, if

$$a,b\in \underset{―}{\mathrm{End}}(C)$$

are homogeneous, then multiplication is composition:

$$ab=a\circ b.$$

The differential satisfies

$$d(ab)=d(a)b+(-1{)}^{|a|}ad(b).$$

This is exactly the graded Leibniz rule for enriched composition.

Thus the dg algebra of endomorphisms is not an extra structure placed on top of a chain complex. It is forced by enrichment:

$$\underset{―}{\mathrm{End}}(C)$$

is the endomorphism monoid object of $C$$C$ inside the dg category $\mathrm{Ch}(\mathcal{C})$$\operatorname{Ch}(\mathcal C)$.

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Summary

The tensor product and Hom complex are not isolated constructions.

The tensor product comes from direct-sum totalization:

$$X{\otimes }_{R}Y={\mathrm{Tot}}^{\oplus }(X_p{\otimes }_R{Y}_q).$$

The Hom complex comes from product totalization:

$$\underset{―}{\mathrm{Hom}}(C,D)={\mathrm{Tot}}^{\mathrm{\Pi }}({\mathrm{Hom}}_{\mathcal{C}}(C_{-p},D_q)).$$

Together they give the closed monoidal structure on $\mathrm{Ch}(R)$$\operatorname{Ch}(R)$:

$$-{\otimes }_{R}Y⊣{\underset{―}{\mathrm{Hom}}}_R(Y,-).$$

The Hom complex also gives the enriched Hom objects for the dg category

$$\mathrm{Ch}(\mathcal{C}).$$

Composition is not merely a graded bilinear operation. It is a chain map:

$$\mathrm{Map}(D,E)\otimes \mathrm{Map}(C,D)\to \mathrm{Map}(C,E).$$

Taking $Z_0$$Z_0$ recovers ordinary composition of chain maps.

Taking $H_0$$H_0$ gives composition in the homotopy category.

Finally, the endomorphism object

$$\underset{―}{\mathrm{End}}(C)$$

is automatically a monoid object in $\mathrm{Ch}(K)$$\operatorname{Ch}(K)$, hence a dg algebra.

Thus the elementary Hom differential

$$(df{)}_i=d_D{f}_i-(-1{)}^n{f}_{i-1}{d}_C$$

is the visible trace of a larger structure:

$$\text{bifunctors}⟶\text{signed double complexes}⟶\text{total complexes}⟶\text{dg enrichment}.$$