Homological Algebra

If $P$ is projective, then this is the special case of Definition 3.2.1 in which $\beta ={\mathrm{id}}_B$ and $g=p$. Conversely, suppose we have an epimorphism $g$ and a morphism $\beta$ as in Definition 3.2.1. We then consider the pullback diagram

Now $g$ is an epimorphism, and it follows that $p_1$ is an epimorphism. By assumption, $p_1$ has a splitting map $u$. We set $\alpha =p_2\circ u$. Then

□

Suppose $P$ is projective. Find a generating set of $P$, and let $F$ be the free left $R$-module on this set. Then we have an epimorphism $F\to P$ defined on this basis. Since this is split, we obtain a direct sum decomposition.

On the other hand, suppose we can write some $P$ as a direct summand of a free module $F$, i.e., $\mathit{𝐹\cong 𝑃}\oplus M$ for some $M$. Using Lemma 3.2.2, suppose we have an exact sequence

Consider the commutative diagram

where the maps $p$, $\iota$ arise from the direct sum decomposition and $q$ exists by the projectivity of $F$. We take $\beta =q\circ \iota$. □

By the classification of finitely generated modules over a principal ideal domain, it suffices for finitely generated $R$-modules to show that any $R$-module of the form

for nonzero and nonunit $a_1,a_2,\dots ,a_n\in R$ is not projective. Consider the obvious quotient map $R^n\to A$. That it splits means that each $R\to R∕(a_i)$ splits. Then $\mathit{𝑅\cong }(a_i)\oplus (x)$ for some $x\in R$, which means that $R$ is free of rank $2$ over itself, which is impossible (e.g., by the classification theorem).

The general case is left as an exercise. □

Suppose that the functor is exact. Then for any epimorphism $g:B\to P$ we have an epimorphism

and any inverse image $t$ of ${\mathrm{id}}_P$ is the desired splitting map of $g$.

On the other hand, suppose that $P$ is projective. Consider an exact sequence

Then we have a diagram

That this is a complex is immediate. Surjectivity of $h_P(g)$ follows immediately from the definition of a projective module. Finally, let $h\in \ker h_P(g)$, so $h:P\to \ker g$. Then $A\to \ker g$ is an epimorphism, and we have by projectivity of $P$ a map $j:P\to A$ with with $f\circ j=h$, i.e., $h_P(f)(j)=h$. □

Let $P=(P_i,d_i)$ and $Q=(Q_i,d_i^{\prime })$, and let $𝜖$ and ${𝜖}^{\prime }$ denote the respective augmentation maps. Then $g\circ 𝜖:P_0\to B$. Since ${𝜖}^{\prime }$ is an epimorphism, we have a map $f_0:P_0\to Q_0$ lifting $g\circ 𝜖$. Now $f_0$ induces a map

and since $\mathrm{im}{d}_1=\ker 𝜖$ and $\ker d_1^{\prime }=\ker {𝜖}^{\prime }$, we have an epimorphism $Q_1\to \ker d_1^{\prime }$, and we again use projectivity, this time of $Q_1$, to lift ${\overline{f}}_0\circ d_1$ to a map $f_1$ as in the diagram. We continue in this manner to obtain $f_{\cdot }$.

Now, for uniqueness up to chain homotopy, it suffices to show that if $g=0$, then $f_{\cdot }$ is chain homotopic to zero. Well, $d_0^{\prime }\circ f_0=g\circ d_0=0$, so $f_0(P_0)\subseteq \mathrm{im}{d}_1^{\prime }$. By projectivity of $P_0$, we have $s_0:P_0\to Q_1$ with

where we have set $s_i=0$ for $i<0$ (and $d_i=0$ for $i\le 0$). Now $h_1=f_1-s_0\circ d_1$ satisfies

so $\mathrm{im}{h}_1\subseteq \mathrm{im}{d}_2^{\prime }$. Thus, we have $s_1:P_1\to Q_1$ lifting $h_1$, i.e., so that

as desired. We continue in this fashion to obtain all $s_i$. □

Choose a lift $t_0$ of ${𝜖}^C$ to $P_0^C\to B$, and let

Then ${𝜖}^B$ is clearly surjective, and we have the desired commutativity of the “first two” squares. Next, letting $d_{\cdot }^X$ denote the boundary maps with $X=A$, $C$, we may define the boundary map $d_1^B$ for $B$ similarly. That is, consider a lift of $d_1^C:P_1^C\to \ker {𝜖}^C$ to a map $t_1:P_1^B\to \ker {𝜖}^B$, and define

Then $d_1^B$ maps onto $\ker d_1^B$ and makes the next two squares commute. We then continue in this fashion. □

Let $P_{\cdot }$ be a split exact sequence of projectives in $𝒞$. In other words, we may write each $P_0=Q_0$ and $P_i=Q_i\oplus Q_{i-1}$ for $i\ge 1$, where $Q_i$ is a projective object in $𝒞$, and the morphism $P_i\to P_{i-1}$ is simply the composition of the projection $P_i\to Q_{i-1}$ with the inclusion $Q_{i-1}\to P_i$. Suppose that ${\pi }_{\cdot }:A_{\cdot }\to P_{\cdot }$ is a epimorphism of complexes. Since $Q_i$ is projective, there exists a splitting $s_i:Q_i\to A_i$ of the composition of ${\pi }_i$ with projection to $Q_i$. Then

is a splitting of ${\pi }_i$. Since $t_{\cdot }$ is a morphism of complexes, it is a splitting of ${\pi }_{\cdot }$. □

Let $P$ be a projective object in $𝒞$. Let $f:A\to B$ be a epimorphism in $𝒟$. We must show that $h_{F(P)}(f):F(A)\to F(B)$ is an epimorphism. Note that we have a commutative diagram

Exactness of $G$ tells us that $G(f)$ is an epimorphism, and the upper horizontal map is then an epimorphism by the projectivity of $P$, hence the result. □

By Proposition 3.2.12, the identity morphism on $A$ extends to a map $P_{\cdot }\to Q_{\cdot }$ of chain complexes. Note also that we have a map $Q_{\cdot }\to P_{\cdot }$, and again by Proposition 3.2.12, both compositions are chain-homotopic to the identity. By Lemma 2.7.15, we therefore have that the induced maps $H_i(F(P_{\cdot }))\to H_i(F(Q_{\cdot }))$ are inverse to each other. □

Since $F$ is right exact, the sequence

is exact. Hence, we have

□

Consider the projective resolution that is $P$ in degree zero and $0$ elsewhere, where the augmentation map $P\to P$ is the identity. This has the desired homology. □

The unique morphism $L_{i}F(f)$ is induced on homology by the morphism of chain complexes given in Proposition 3.2.12. Functoriality follows by canonicality of the map of homology.

To see additivity, note that $L_{i}F(0_A)$ is induced by the zero morphism of chain complexes and hence is is zero map on $L_{i}F(A)$. Similarly $L_{i}F(f+g)$, for $f,g:A\to B$, can be given by the sum of the induced maps on chain complexes, hence is given by the sum of the maps on homology. □

By the Horseshoe lemma, we have a projective resolution $P_{\cdot }^X\to X$ for $X=A$, $B$, $C$ fitting in a diagram (3.2.1). Now, applying $F$ to the resolutions, we have split exact sequences

for each $i$. The resulting exact sequence of complexes (which need not be split) yields a long exact sequence in homology

as desired.

It remains to check naturality. Consider a morphism of short exact sequences

By Proposition 3.2.12, can extend $q^A$ and $q^C$ to maps of complexes $q_{\cdot }^A:P_{\cdot }^A\to P_{\cdot }^{A^{\prime }}$ and $q_{\cdot }^C:P_{\cdot }^C\to P_{\cdot }^{C^{\prime }}$. Suppose we have constructed $P_{\cdot }^B$ and $P_{\cdot }^{B^{\prime }}$ via the Horseshoe lemma. We fit this all into a commutative diagram

Diagram description: Maps between horseshoe resolutions The structural squares and triangles displayed here commute. Objects, listed by row and column: Row 1, from left to right: column 1: 0; column 3: P subscript (dot) superscript (A); column 5: P subscript (dot) superscript (B); column 7: P subscript (dot) superscript (C); column 9: 0. Row 2, from left to right: column 2: 0; column 4: A; column 6: B; column 8: C; column 10: 0. Row 3, from left to right: column 1: 0; column 3: P subscript (dot) superscript (A prime); column 5: P subscript (dot) superscript (B prime); column 7: P subscript (dot) superscript (C prime); column 9: 0. Row 4, from left to right: column 2: 0; column 4: A prime; column 6: B prime; column 8: C prime; column 10: 0. Arrows and lines: An arrow from 0 (row 1, column 1) to P subscript (dot) superscript (A), without a label. An arrow from P subscript (dot) superscript (A) to P subscript (dot) superscript (B), labelled iota subscript (dot). An arrow from P subscript (dot) superscript (A) to A, labelled epsilon superscript (A). An arrow from P subscript (dot) superscript (A) to P subscript (dot) superscript (A prime), labelled q subscript (dot) superscript (A). An arrow from P subscript (dot) superscript (B) to P subscript (dot) superscript (C), labelled p subscript (dot). An arrow from P subscript (dot) superscript (B) to B, without a label. An arrow from P subscript (dot) superscript (C) to C, without a label. An arrow from P subscript (dot) superscript (C) to P subscript (dot) superscript (C prime), labelled q subscript (dot) superscript (C). An arrow from P subscript (dot) superscript (C) to 0 (row 1, column 9), without a label. An arrow from 0 (row 2, column 2) to A, without a label. An arrow from A to B, labelled f. An arrow from A to A prime, without a label. An arrow from B to C, labelled g. An arrow from B to B prime, without a label. An arrow from C to C prime, labelled q superscript (C). An arrow from C to 0 (row 2, column 10), without a label. An arrow from 0 (row 3, column 1) to P subscript (dot) superscript (A prime), without a label. An arrow from P subscript (dot) superscript (A prime) to P subscript (dot) superscript (B prime), labelled iota prime subscript (dot). An arrow from P subscript (dot) superscript (A prime) to A prime, without a label. An arrow from P subscript (dot) superscript (B prime) to B prime, without a label. An arrow from P subscript (dot) superscript (B prime) to P subscript (dot) superscript (C prime), labelled p prime subscript (dot). An arrow from P subscript (dot) superscript (C prime) to C prime, without a label. An arrow from P subscript (dot) superscript (C prime) to 0 (row 3, column 9), without a label. An arrow from 0 (row 4, column 2) to A prime, without a label. An arrow from A prime to B prime, labelled f prime. An arrow from B prime to C prime, labelled g prime. An arrow from C prime to 0 (row 4, column 10), without a label.

(3.3.1)

We also have splitting maps $j_i:P_i^C\to P_i^B$ and $k_i:P_i^B\to P_i^A$ for each $i$ (and, similarly, maps $j_i^{\prime }$and $k_i^{\prime }$). For each $X$, let us denote the augmentation map by ${𝜖}^X$.

We must define a map $q_{\cdot }^B:P_{\cdot }^B\to P_{\cdot }^{B^{\prime }}$ making the entire diagram (3.3.1) commute. We first note that

Hence, there exists a map ${\beta }_0:P_0^B\to A^{\prime }$ with

Since ${𝜖}^{A^{\prime }}$is an epimorphism, we may choose ${\alpha }_0:P_0^B\to P_0^{A^{\prime }}$ with ${𝜖}^{A^{\prime }}\circ {\alpha }_0={\beta }_0$. Now set

The trickiest check of commutativity is that ${𝜖}^{B^{\prime }}\circ q_0^B=q^B\circ {𝜖}^B$. We write this mess out:

The other $q_i^B$ are defined similarly. For instance, one can see there exists a map ${\beta }_1:P_1^C\to P_0^{A^{\prime }}$ such that

and we set

□

Note that we have an exact sequence

so $F(A)\cong H_0(F(C_{\cdot }))$. Set $K_0=\ker {𝜖}^C$. We then have an exact sequence

which yields

We also have isomorphisms $L_{i}F(A)\cong L_{i-1}F(K_0)$ for each $i\ge 2$.

For $i\ge 1$, set $K_i=\ker d_i^C\cong \mathrm{im}{d}_{i+1}^C$. The exact sequences

then yield isomorphisms used in the following for $i\ge 2$:

□

Suppose that $(G_i,{\delta }_i^{\prime })$ is another $\delta$-functor and that we have a natural transformation $G_0⇝F_0$. Let $A\in \mathrm{Obj}(𝒞)$ and let $\pi :P\to A$ be an epimorphism with $P$ projective. Let $K=\ker \pi$. Let $i\ge 1$, and suppose that we have constructed a natural transformation $G_{i-1}⇝F_{i-1}$. Since $F_i(P)=0$ is projective, we have a commutative diagram

The morphism $G_i(A)\to F_i(A)$ is the unique map which makes the diagram commute.

Now let $f:A\to B$ be a morphism in $𝒞$. We create a diagram as follows:

by taking $P^{\prime }$ to be projective, $P$ to be any projective with an epimorphism to the pullback of the diagram $P^{\prime }\to B\leftarrow A$, and $K$ and $K^{\prime }$ to be the relevant kernels. We then have a diagram

We need only see that the leftmost square commutes, but this follows easily from a diagram chase and the fact that the two horizontal maps on the frontmost square are monomorphisms.

Hence, we have constructed a sequence of natural transformations $G_i⇝F_i$. It remains only to see that these form a morphism of $\delta$-functors. This being an inductive argument of the above sort, we leave it to the reader. □

Let $D$ be injective, and take $d\in D$. Then there exists a group homomorphism $\varphi :\mathbb{Z}\to D$ with $1\mapsto d$. We also have the multiplication-by-$n$ map on $\mathbb{Z}$, which is injective. By injectivity of $D$, we have a map $𝜃:\mathbb{Z}\to D$ with $\varphi =\mathit{𝑛𝜃}$. Then $d=\mathit{𝑛𝜃}(1)$, so $D$ is divisible.

Conversely, let $D$ be divisible. By Lemma 3.4.3, it suffices to show that every homomorphism $\varphi :\mathit{𝑛\mathbb{Z}}\to D$ with $n\ge 1$ extends to a homomorphism $𝜃:\mathbb{Z}\to D$. Such a $\varphi$ is determined by $d=\varphi (n)$. Let $d^{\prime }\in D$ be such that $n{d}^{\prime }=d$. Set $𝜃(1)=d^{\prime }$. □

First take the case that $R=\mathbb{Z}$. Let $A$ be an abelian group, and write it as a quotient of a free abelian group

for some indexing set $J$ and submodule $T$ of ${\oplus }_{j\in J}\mathbb{Z}$. Then we may embed $A$ in