Evaluation, Disks, and Projective Chain Complexes

 

Exercise 2.2.1

Let $\mathcal{A}$$\mathcal A$ be an abelian category. We use the homological convention

$$\cdots ⟶X_{n+1}\stackrel{d_{n+1}}{\to }{X}_n\stackrel{d_n}{\to }{X}_{n-1}⟶\cdots .$$

For a complex $X$$X$, write

$$Z_n(X)=\ker (d_n:X_n\to X_{n-1}).$$

We prove that a complex $P\in \mathrm{Ch}(\mathcal{A})$$P\in \operatorname{Ch}(\mathcal A)$ is projective if and only if $P$$P$ is split exact and each $P_n$$P_n$ is projective in $\mathcal{A}$$\mathcal A$.

First recall that $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$ is abelian, with kernels and cokernels computed degreewise. Hence a chain map

$$f:X\to Y$$

is an epimorphism in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$ if and only if every

$$f_n:X_n\to Y_n$$

is an epimorphism in $\mathcal{A}$$\mathcal A$.

We shall use the following standard principle. Let

$$L:\mathcal{A}\rightleftarrows \mathcal{B}:R$$

be an additive adjunction between abelian categories,

$$L⊣R.$$

If $R$$R$ is exact, then $L$$L$ preserves projective objects. Indeed, if $P$$P$ is projective in $\mathcal{A}$$\mathcal A$, then

$${\mathrm{Hom}}_{\mathcal{A}}(P,-)$$

is exact. By adjunction,

$${\mathrm{Hom}}_{\mathcal{B}}(L(P),-)\cong {\mathrm{Hom}}_{\mathcal{A}}(P,R(-)).$$

The right-hand side is a composite of exact functors, hence is exact. Therefore $L(P)$$L(P)$ is projective in $\mathcal{B}$$\mathcal B$.

For $A\in \mathcal{A}$$A\in\mathcal A$, define the disk complex $D^n(A)$$D^n(A)$ by

$$D^n(A):\cdots \to 0\to A\stackrel{1_A}{\to }A\to 0\to \cdots ,$$

where the two copies of $A$$A$ lie in degrees $n$$n$ and $n-1$$n-1$. Thus

$$\begin{array}{c}{D}^n(A{)}_k=\{\begin{array}{ll}A,& k=n,\\ A,& k=n-1,\\ 0,& \text{otherwise},\end{array}\end{array}$$

and the only nonzero differential is

$$d_n^{D^n(A)}=1_A.$$

For each $n$$n$, the evaluation functor

$${\mathrm{ev}}_n:\mathrm{Ch}(\mathcal{A})\to \mathcal{A},X\mapsto X_n$$

has both a left adjoint and a right adjoint:

$$D^n(-)⊣{\mathrm{ev}}_n⊣D^{n+1}(-).$$

Indeed,

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(D^n(A),X)\cong {\mathrm{Hom}}_{\mathcal{A}}(A,X_n),$$

because a chain map $D^n(A)\to X$$D^n(A)\to X$ is determined by its degree $n$$n$ component $A\to X_n$$A\to X_n$, and the degree $n-1$$n-1$ component is forced to be its composite with $d_n^X$$d_n^X$.

Similarly,

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(X,D^{n+1}(A))\cong {\mathrm{Hom}}_{\mathcal{A}}(X_n,A),$$

because a chain map $X\to D^{n+1}(A)$$X\to D^{n+1}(A)$ is determined by its degree $n$$n$ component $X_n\to A$$X_n\to A$, and the degree $n+1$$n+1$ component is forced to be its composite with $d_{n+1}^X$$d_{n+1}^X$.

The functor ${\mathrm{ev}}_n$$\operatorname{ev}_n$ is exact, because exactness in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$ is degreewise. The functor $D^n(-)$$D^n(-)$ is also exact, since it places the same short exact sequence in two degrees and zero elsewhere.

Therefore,

$$Q\text{ projective in }\mathcal{A}⟹D^n(Q)\text{ projective in }\mathrm{Ch}(\mathcal{A}).$$

Also, using

$${\mathrm{ev}}_n⊣D^{n+1}(-)$$

and the exactness of $D^{n+1}(-)$$D^{n+1}(-)$, we get

$$P\text{ projective in }\mathrm{Ch}(\mathcal{A})⟹P_n\text{ projective in }\mathcal{A}.$$

Indeed,

$${\mathrm{Hom}}_{\mathcal{A}}(P_n,-)\cong {\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(P,D^{n+1}(-)),$$

and the right-hand side is exact if $P$$P$ is projective.

Now assume that $P$$P$ is projective in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$. We have already shown that every $P_n$$P_n$ is projective in $\mathcal{A}$$\mathcal A$. It remains to show that $P$$P$ is split exact.

The shift $[-1]$$[-1]$ is defined by

$$P[-1{]}_n=P_{n-1},d_n^{P[-1]}=-d_{n-1}^P.$$

Since shift is an exact autoequivalence, it preserves projective objects. Hence $P[-1]$$P[-1]$ is projective.

Consider the mapping cone of the identity map $1_P:P\to P$$1_P:P\to P$. We use the convention

$$\mathrm{Cone}(1_P{)}_n=P_n\oplus P_{n-1},$$

with differential

$$d_n^{\mathrm{Cone}}:P_n\oplus P_{n-1}\to P_{n-1}\oplus P_{n-2}$$

given by

$$d_n^{\mathrm{Cone}}(b,a)=(d_n^{P}b+a,-d_{n-1}^{P}a).$$

Equivalently,

$$\begin{array}{c}{d}_n^{\mathrm{Cone}}=\left[\begin{array}{cc}{d}_n^P& 1_{P_{n-1}}\\ 0& -d_{n-1}^P\end{array}\right].\end{array}$$

There is a short exact sequence

$$0\to P\stackrel{j}{\to }\mathrm{Cone}(1_P)\stackrel{q}{\to }P[-1]\to 0,$$

where

$$j_n(b)=(b,0),q_n(b,a)=a.$$

The map $j$$j$ is a chain map because

$$d_n^{\mathrm{Cone}}{j}_n(b)=d_n^{\mathrm{Cone}}(b,0)=(d_n^{P}b,0)=j_{n-1}{d}_n^P(b).$$

The map $q$$q$ is a chain map because

$$q_{n-1}{d}_n^{\mathrm{Cone}}(b,a)=q_{n-1}(d_n^{P}b+a,-d_{n-1}^{P}a)=-d_{n-1}^{P}a,$$

while

$$d_n^{P[-1]}{q}_n(b,a)=d_n^{P[-1]}(a)=-d_{n-1}^{P}a.$$

Degreewise, the sequence

$$0\to P_n\to P_n\oplus P_{n-1}\to P_{n-1}\to 0$$

is split exact, so the sequence is exact in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$.

Since $P[-1]$$P[-1]$ is projective, the epimorphism

$$q:\mathrm{Cone}(1_P)\twoheadrightarrow P[-1]$$

has a chain map section

$$s:P[-1]\to \mathrm{Cone}(1_P)$$

such that

$$qs=1_{P[-1]}.$$

At degree $n$$n$,

$$s_n:P_{n-1}\to P_n\oplus P_{n-1}.$$

Since $q_n{s}_n=1_{P_{n-1}}$$q_ns_n=1_{P_{n-1}}$, the section must have the form

$$s_n(a)=(h_{n-1}(a),a),$$

for some morphism

$$h_{n-1}:P_{n-1}\to P_n.$$

Now impose the chain map condition

$$d_n^{\mathrm{Cone}}{s}_n=s_{n-1}{d}_n^{P[-1]}.$$

For $a\in P_{n-1}$$a\in P_{n-1}$, the left-hand side is

$$d_n^{\mathrm{Cone}}{s}_n(a)=d_n^{\mathrm{Cone}}(h_{n-1}(a),a)=(d_n^P{h}_{n-1}(a)+a,-d_{n-1}^{P}a).$$

The right-hand side is

$$s_{n-1}{d}_n^{P[-1]}(a)=s_{n-1}(-d_{n-1}^{P}a)=(-h_{n-2}{d}_{n-1}^P(a),-d_{n-1}^P(a)).$$

Equating first components gives

$$d_n^P{h}_{n-1}+1_{P_{n-1}}=-h_{n-2}{d}_{n-1}^P.$$

Hence

$$d_n^P{h}_{n-1}+h_{n-2}{d}_{n-1}^P=-1_{P_{n-1}}.$$

Renaming $m=n-1$$m=n-1$, we get

$$d_{m+1}^P{h}_m+h_{m-1}{d}_m^P=-1_{P_m}.$$

Set

$$k_m=-h_m:P_m\to P_{m+1}.$$

Then

$$d_{m+1}^P{k}_m+k_{m-1}{d}_m^P=1_{P_m}.$$

Thus $P$$P$ is contractible.

We now show that contractible implies split exact. Since

$$d_m{d}_{m+1}=0,$$

the differential $d_{m+1}:P_{m+1}\to P_m$$d_{m+1}:P_{m+1}\to P_m$ factors through $Z_m(P)$$Z_m(P)$:

$$P_{m+1}\stackrel{{\bar{d}}_{m+1}}{\to }{Z}_m(P)\stackrel{i_m}{\to }{P}_m,$$

where $i_m:Z_m(P)\to P_m$$i_m:Z_m(P)\to P_m$ is the kernel inclusion and

$$i_m{\bar{d}}_{m+1}=d_{m+1}.$$

Define

$${\sigma }_m:Z_m(P)\to P_{m+1}$$

by

$${\sigma }_m=k_m{i}_m.$$

Then

$$i_m{\bar{d}}_{m+1}{\sigma }_m=d_{m+1}{k}_m{i}_m.$$

Using

$$d_{m+1}{k}_m+k_{m-1}{d}_m=1_{P_m},$$

we get

$$d_{m+1}{k}_m{i}_m=(1_{P_m}-k_{m-1}{d}_m)i_m.$$

But $d_m{i}_m=0$$d_mi_m=0$, so

$$d_{m+1}{k}_m{i}_m=i_m.$$

Therefore

$$i_m{\bar{d}}_{m+1}{\sigma }_m=i_m.$$

Since $i_m$$i_m$ is monic,

$${\bar{d}}_{m+1}{\sigma }_m=1_{Z_m(P)}.$$

Thus

$${\bar{d}}_{m+1}:P_{m+1}\to Z_m(P)$$

is a split epimorphism. Its kernel is $Z_{m+1}(P)$$Z_{m+1}(P)$, so we get a split short exact sequence

$$0\to Z_{m+1}(P)\to P_{m+1}\stackrel{{\bar{d}}_{m+1}}{\to }{Z}_m(P)\to 0.$$

This holds for every $m$$m$, hence $P$$P$ is split exact.

We have shown that if $P$$P$ is projective in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$, then $P$$P$ is split exact and every $P_n$$P_n$ is projective in $\mathcal{A}$$\mathcal A$.

Conversely, suppose that $P$$P$ is split exact and every $P_n$$P_n$ is projective in $\mathcal{A}$$\mathcal A$.

For every $n$$n$, split exactness gives a split short exact sequence

$$0\to Z_n(P)\to P_n\to Z_{n-1}(P)\to 0.$$

Thus

$$P_n\cong Z_n(P)\oplus Z_{n-1}(P).$$

Since $Z_n(P)$$Z_n(P)$ is a direct summand of the projective object $P_n$$P_n$, it is projective. Hence every $Z_n(P)$$Z_n(P)$ is projective in $\mathcal{A}$$\mathcal A$.

Choose splittings

$$P_n\cong Z_n(P)\oplus Z_{n-1}(P).$$

Under these splittings, the differential has the form

$$d_n:Z_n(P)\oplus Z_{n-1}(P)\to Z_{n-1}(P)\oplus Z_{n-2}(P),$$

$$d_n(z,w)=(w,0).$$

Indeed, the first summand $Z_n(P)$$Z_n(P)$ lies in the kernel of $d_n$$d_n$, while the chosen section of $Z_{n-1}(P)\to P_n$$Z_{n-1}(P)\to P_n$ maps by $d_n$$d_n$ identically onto $Z_{n-1}(P)$$Z_{n-1}(P)$.

Therefore $P$$P$ is the locally finite coproduct of disk complexes

$$P\cong \coprod _{n\in \mathbb{Z}}^{\mathrm{loc}.\mathrm{fin}.}{D}^{n+1}(Z_n(P)).$$

Here "locally finite" means that in each degree only finitely many summands are nonzero. In degree $m$$m$, only the disks

$$D^{m+1}(Z_m(P))\text{and}{D}^m(Z_{m-1}(P))$$

contribute.

Now let $X$$X$ be any chain complex. Since $\mathrm{Hom}$$\operatorname{Hom}$ out of a coproduct is a product,

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(P,X)\cong \prod _{n\in \mathbb{Z}}{\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(D^{n+1}(Z_n(P)),X).$$

By the disk adjunction,

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(D^{n+1}(Z_n(P)),X)\cong {\mathrm{Hom}}_{\mathcal{A}}(Z_n(P),X_{n+1}).$$

Hence

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(P,X)\cong \prod _{n\in \mathbb{Z}}{\mathrm{Hom}}_{\mathcal{A}}(Z_n(P),X_{n+1}).$$

Now let

$$e:X\twoheadrightarrow Y$$

be an epimorphism in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$. Since epimorphisms of complexes are degreewise epimorphisms, each

$$e_{n+1}:X_{n+1}\twoheadrightarrow Y_{n+1}$$

is an epimorphism in $\mathcal{A}$$\mathcal A$.

Since $Z_n(P)$$Z_n(P)$ is projective, the induced map

$${\mathrm{Hom}}_{\mathcal{A}}(Z_n(P),X_{n+1})\to {\mathrm{Hom}}_{\mathcal{A}}(Z_n(P),Y_{n+1})$$

is surjective for every $n$$n$.

Taking products over all $n$$n$, we get a surjection

$$\prod _n{\mathrm{Hom}}_{\mathcal{A}}(Z_n(P),X_{n+1})\to \prod _n{\mathrm{Hom}}_{\mathcal{A}}(Z_n(P),Y_{n+1}).$$

Under the Hom-identifications above, this is exactly the map

$${\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(P,X)\to {\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{A})}(P,Y)$$

induced by $e:X\to Y$$e:X\to Y$. Therefore every chain map $P\to Y$$P\to Y$ lifts along every epimorphism $X\twoheadrightarrow Y$$X\twoheadrightarrow Y$. Hence $P$$P$ is projective in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$.

Thus

$$P\text{ is projective in }\mathrm{Ch}(\mathcal{A})⟺P\text{ is split exact and every }{P}_n\text{ is projective in }\mathcal{A}.$$

Exercise 2.2.2

Assume that $\mathcal{A}$$\mathcal A$ has enough projectives. We prove that $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$ has enough projectives.

Let

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

be any chain complex:

$$\cdots \to C_{n+1}\stackrel{d_{n+1}^C}{\to }{C}_n\stackrel{d_n^C}{\to }{C}_{n-1}\to \cdots .$$

Since $\mathcal{A}$$\mathcal A$ has enough projectives, for every $n$$n$ choose a projective object $Q_n$$Q_n$ and an epimorphism

$${\pi }_n:Q_n\twoheadrightarrow C_n.$$

Define a complex $P$$P$ by

$$P_n=Q_n\oplus Q_{n+1}.$$

Define the differential

$$d_n^P:P_n\to P_{n-1}$$

by

$$d_n^P(q_n,q_{n+1})=(0,q_n).$$

Equivalently,

$$d_n^P:Q_n\oplus Q_{n+1}\to Q_{n-1}\oplus Q_n$$

is the matrix

$$\begin{array}{c}{d}_n^P=\left[\begin{array}{cc}0& 0\\ 1_{Q_n}& 0\end{array}\right].\end{array}$$

Then

$$d_{n-1}^P{d}_n^P(q_n,q_{n+1})=d_{n-1}^P(0,q_n)=(0,0).$$

Hence $P$$P$ is a chain complex.

We now check that $P$$P$ is split exact. Since

$$d_n^P(q_n,q_{n+1})=(0,q_n),$$

we have

$$Z_n(P)=0\oplus Q_{n+1}.$$

Also,

$$d_{n+1}^P(q_{n+1},q_{n+2})=(0,q_{n+1}),$$

so

$$\mathrm{im}(d_{n+1}^P)=0\oplus Q_{n+1}=Z_n(P).$$

The induced map

$$P_{n+1}=Q_{n+1}\oplus Q_{n+2}\to Z_n(P)\cong Q_{n+1}$$

is the projection onto $Q_{n+1}$$Q_{n+1}$, and it has the section

$$Q_{n+1}\to Q_{n+1}\oplus Q_{n+2},q_{n+1}\mapsto (q_{n+1},0).$$

Therefore $P$$P$ is split exact.

Each

$$P_n=Q_n\oplus Q_{n+1}$$

is projective in $\mathcal{A}$$\mathcal A$, since finite direct sums of projective objects are projective. By Exercise 2.2.1, $P$$P$ is projective in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$.

It remains to construct an epimorphism

$$P\twoheadrightarrow C.$$

Define a chain map

$$F:P\to C$$

degreewise by

$$F_n:Q_n\oplus Q_{n+1}\to C_n,$$

$$F_n(q_n,q_{n+1})={\pi }_n(q_n)+d_{n+1}^C{\pi }_{n+1}(q_{n+1}).$$

Equivalently,

$$F_n=\left[\begin{array}{cc}{\pi }_n& d_{n+1}^C{\pi }_{n+1}\end{array}\right].$$

We check that $F$$F$ is a chain map. First,

$$d_n^C{F}_n(q_n,q_{n+1})=d_n^C{\pi }_n(q_n)+d_n^C{d}_{n+1}^C{\pi }_{n+1}(q_{n+1}).$$

Since $C$$C$ is a chain complex,

$$d_n^C{d}_{n+1}^C=0.$$

Therefore

$$d_n^C{F}_n(q_n,q_{n+1})=d_n^C{\pi }_n(q_n).$$

On the other hand,

$$F_{n-1}{d}_n^P(q_n,q_{n+1})=F_{n-1}(0,q_n).$$

By definition of $F_{n-1}$$F_{n-1}$,

$$F_{n-1}(0,q_n)=d_n^C{\pi }_n(q_n).$$

Hence

$$d_n^C{F}_n=F_{n-1}{d}_n^P.$$

So $F$$F$ is a chain map.

Finally, $F_n$$F_n$ is an epimorphism for every $n$$n$, because its restriction to the first summand is

$${\pi }_n:Q_n\twoheadrightarrow C_n.$$

Thus $F$$F$ is degreewise epi, hence epi in $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$.

We have constructed a projective complex $P$$P$ and an epimorphism

$$P\twoheadrightarrow C.$$

Since $C$$C$ was arbitrary, $\mathrm{Ch}(\mathcal{A})$$\operatorname{Ch}(\mathcal A)$ has enough projectives.