## An Introduction to Tor: Definitions and Properties.

Last update: 09 March 2012

## Introduction

Homology is one of the most important concepts of modern mathematics. Its importance comes from the fact that homology of a space is invariant under continuous transformations of the space.

Homological algebra develops the algebraic tools used to study homology. Tensor products, Hom, Tor, and Ext are basic functors of homological algebra. This paper introduces the functor Tor and its properties. Tor is defined precisely and several properties are derived.

In order to make the exposition of Tor complete, a short section with preliminary definitions and properties of homology is included. This presentation will serve to reacquaint the reader with basic algebraic topology. This section also provides a concise reference to the basics of algebraic topology. It is not meant to serve as an introduction to basic algebra and algebraic topology, but merely to make the treatment of Tor precise and complete. Definitions and properties of tensor products are included in this section, as they play a central role in the definition of Tor.

Tensor product and projective resolutions are the primary tools for defining Tor. The definition of projective resolutions is done meticulously in order to provide a solid basis for introducing Tor precisely.

Given the definition of Tor, several properties can be derived easily. These properties will be introduced and should provide a good foundation for working with Tor in further study of algebraic topology.

## Basic Algebra and Algebraic Topology

We begin with the following treatment of modules and homology. It should provide a sufficient basis for subsequent proofs.

Throughout this paper, we shall let $R$ represent a commutative ring with a unit.

### Modules

The following definitions of $R-$modules and $R-$maps are the standard ones. Since the intention here is only to provide a solid algebraic basis for introducing Tor, we will not give proofs of the theorems and propositions in this section. A more complete treatment is given by Atiyah and MacDonald [AM].

An $R-$module $M$ is an abelian group with a map $f:R×M\to M$ such that if $f\left(r,m\right)$ is denoted by $rm,$ then $\left({r}_{1}{r}_{2}\right)m={r}_{1}\left({r}_{2}m\right)$ and $1m=m.$ We usually denote $R-$modules simply as modules.

Let $M,N$ be $R-$modules. An $R-$map $f:M\to N$ is a homomorphism of abelian groups such that $f\left(rm\right)=rf\left(m\right).$ We shall assume that all maps between modules are $R-$maps.

If ${\left({M}_{i}\right)}_{i\in I}$ is any family of $R-$modules, their direct sum $\underset{i\in I}{⨁}{M}_{i}$ is the set of all families $\left({m}_{i}\right)$ such that ${m}_{i}\in {M}_{i}$ and almost all the ${m}_{i}$ are zero. Addition and multiplication are defined by:

We now define tensor products. Let and $P$ be $R-$modules.

A bilinear map $h:M×N\to P$ is a map such that and $h\left(rm,n\right)=h\left(m,rn\right)$ for $r\in R.$

The tensor product $M\otimes N$ is an $R-$module with a bilinear map $h:M×N\to M\otimes N$ such that given any bilinear map $g:M×N\to P$ there is a unique map $f:M\otimes N\to P$ such that $g=fh.$ Tensor product can be defined equivalently by and the relations The proof that these definitions are equivalent can be found in most texts that introduce tensor products. See Munkres [Mu].

The following proposition summarizes the important properties of tensor product.

Let be $R-$modules.

1. $M\otimes N=N\otimes M$
2. $\left(M\otimes N\right)\otimes P=M\otimes \left(N\otimes P\right)=M\otimes N\otimes P$
3. $\left(M\oplus N\right)\otimes P=\left(M\otimes P\right)\oplus \left(N\otimes P\right)$
4. $R\otimes M=M$

An exact sequence $⋯→Mi+1→pi+1 Mi→pi Mi-1→⋯$ of $R-$modules is a sequence of modules $\left\{{M}_{i}\right\}$ and $R-$maps ${p}_{i}:{M}_{i}\to {M}_{i-1}$ such that $\mathrm{ker}{p}_{i}=\mathrm{im}{p}_{i+1}$ for all $i.$

An exact sequence of the form $0→M'→M→M''→0,$ is a short exact sequence. A short exact sequence is a split short exact sequence if there exists a map $p:M\text{'}\oplus M\text{'}\text{'}\to M$ such that the following diagram commutes when $k$ is given by $k\left(m\text{'}\right)=\left(m\text{'},0\right)$ and $1$ is given by $1\left(m\text{'},m\text{'}\text{'}\right)=m\text{'}\text{'}.$

If $0→0000M'→0000M→0000M''→00000$ is an exact sequence of $R-$modules, and $N$ is an $R-$module, then $M'⊗N→0000M⊗N→0000M''⊗N→00000$ is exact.

If $0→0000M'→0000M→0000M''→00000$ is a split short exact sequence of $R-$modules, and $N$ is an $R-$module, then $0→0000M'⊗N→0000M⊗N→0000M''⊗N→00000$ is split short exact.

The above is a very concise presentation of modules, tensor products, and exact sequences.

### Homology

We treat homology in a completely algebraic manner. Although the algebraic presentation is completely precise, the power of homology lies in its geometrical application and it is usually presented in a geometric context. Munkres' boo [Mu] has a complete development of homology both algebraically and geometrically.

A chain complex is a sequence of $R-$modules ${C}_{q}$ and $R-$maps ${\partial }_{q}:{C}_{q}\to {C}_{q-1}$ such that ${\partial }_{q-1}{\partial }_{q}=0$ for all $q.$

Given a chain complex $C$ the homology of $C$ is given by the family of groups ${H}_{q}\left(C\right)=\frac{\mathrm{ker}{\partial }_{q}}{\mathrm{im}{\partial }_{q+1}}.$

Let $C=\left\{{C}_{q}\right\}$ and $C\text{'}=\left\{{C}_{q}\text{'}\right\}$ be chain complexes where ${\partial }_{q}$ are the $R-$maps on $C$ and ${\partial }_{q}\text{'}$ are the $R-$maps on $C\text{'}.$ A chain map $f:C\to C\text{'}$ is a family of $R-$maps ${f}_{q}:{C}_{q}\to {C}_{q}\text{'}$ such that ${\partial }_{q}\text{'}{f}_{q}={f}_{q-1}{\partial }_{q}$ for all $q.$

A chain map $f:C\to C\text{'}$ induces homomorphisms ${{f}_{q}}_{*}:{H}_{q}\left(C\right)\to {H}_{q}\left(C\text{'}\right)$ for all $q.$

Given chain maps $f,g:C\to C\text{'},$ then a chain homotopy $D$ from $f$ to $g$ is a family of $R-$maps ${D}_{q}:{C}_{q}\to {C}_{q+1}\text{'}$ such that ${\partial }_{q+1}\text{'}{D}_{q}+{D}_{q-1}{\partial }_{q}={f}_{q}-{g}_{q}$ for all $q.$

Two chain complexes $C$ and $C\text{'}$ are chain equivalent if there exists $f:C\to C\text{'}$ and $g:C\text{'}\to C$ such that $gf$ and $fg$ are both chain homotopic to the identity.

A chain complex is acyclic if the chain complex forms an exact sequence of $R-$modules.

These definitions provide the basis of the study of homology. The following results are part of a collection of results known as the Eilenberg-Steenrod axioms. We only present the axioms that we will need for subsequent proofs.

Let be chain maps.

If $i$ is the identity chain map, then ${i}_{*}$ is an isomorphism.

If $h$ and $k$ are chain homotopic, then ${h}_{*}={k}_{*}.$

(kh)* = k*h*.

Long exact sequence theorem. Given a short exact sequence of chain complexes $0→0000C'→iC→jC''→00000,$ (i.e. $0\stackrel{\phantom{0000}}{\to }{C}_{q}\text{'}\stackrel{{i}_{q}}{\to }{C}_{q}\stackrel{{j}_{q}}{\to }{C}_{q}\text{'}\text{'}\stackrel{\phantom{0000}}{\to }0$ is exact for each $q$) then there exists a homomorphism ${{\partial }_{q}}_{*}:{H}_{q}\left(C\text{'}\text{'}\right)\to {H}_{q-1}\left(C\text{'}\right)$ such that the sequence $⋯→0000 Hq(C')→0000 Hq(C)→0000 Hq(C'')→∂q* Hq-1(C')→0000⋯$ is exact.

The above definitions and terms should make the following definition of Tor complete and readable. This exposition should serve to prevent any confusion in terminology.

## Projective Modules

Projective resolutions are the primary tools for defining Tor. We present them here in a precise and complete manner. The result of Theorem 4.6 is the key to the definition of Tor. It is this result that makes Tor a well defined functor.

We begin by defining a projective $R-$module.

Suppose that $f:M\to M\text{'}\text{'}$ is a surjective $R-$map. $P$ is a projective $R-$module if given a map $h:P\to M\text{'}\text{'},$ there exists a map $g:P\to M$ such that $g=fh.$

Conceptually, this definition says that $P$ is a projective $R-$module if the following diagram can be filled in

Note that $M\stackrel{f}{\to }M\text{'}\text{'}\to 0$ is exact.

The following results give some insight into the implications of this definition.

A free $R-$module is projective.

 Proof. Let $P$ be a free $R-$module. Since $P$ is free, there exists a family $\left\{{p}_{i}\right\}$ of elements of $P$ such that any $p\in P$ can be written uniquely as $p=\sum {r}_{i}{p}_{i}$ with ${r}_{i}\in R.$ Let $h:P\to M\text{'}\text{'}$ and $f:M\to M\text{'}\text{'}$ be $R-$maps such that $f$ is surjective. To show that $P$ is projective, we need to construct a $g:P\to M$ such that $h=fg.$ Conceptually, we show that $P$ is projective by filling in the following diagram: $P$ $M$ $M''$ $0$ $g$ $h$ $f$ Since $f$ is surjective, for each ${p}_{i}\in \left\{{p}_{i}\right\}$ there exists at least one element ${m}_{i}\in M$ such that $f\left({m}_{i}\right)=h\left({p}_{i}\right).$ Choose one such ${m}_{i}$ for each ${p}_{i}$ and let $g\left({p}_{i}\right)={m}_{i}.$ Letting $g\left(p\right)=\sum {r}_{i}g\left({p}_{i}\right)$ when $p=\sum {r}_{i}{m}_{i}$ defines $g\left(p\right)$ for all $p\in P.$ $g\left(p\right)$ is well defined since any $p\in P$ can be written as the sum $p=\sum {r}_{i}{p}_{i}$ in only one way. $g$ is an $R-$map since $g\left(rp\right)=r\sum {r}_{i}g\left({p}_{i}\right)=rg\left(p\right)$ for every $p\in P.$ Furthermore, $f\left(g\left(p\right)\right)=f\left(\sum {r}_{i}g\left({p}_{i}\right)\right)=\sum {r}_{i}f\left(g\left({p}_{i}\right)\right)=\sum {r}_{i}h\left({p}_{i}\right)=h\left(p\right)$ since $f$ and $h$ are $R-$maps. Thus, $fg=h$ and $P$ is projective. $\square$

We will use the following lemma in the proof of Proposition 3.4.

Let $M$ be an $R-$module. There exists a free module $F$ and a surjective map $u\in F\to M.$

 Proof. Let ${\left\{{m}_{i}\right\}}_{i\in I}$ be a generating set for $M.$ Let $\underset{I}{⨁}R=F$ be a free module with basis ${\left\{{x}_{i}\right\}}_{i\in I}.$ Then the map $u:F\to M$ given by $u\left(r\right)=\sum {r}_{i}{g}_{i}$ for $r=\sum {r}_{i}{x}_{i}$ is surjective. $\square$

$P$ is projective if and only if $P$ is a direct summand of a free module.

 Proof. Suppose that $P$ is projective. Let $u:\underset{I}{⨁}R\to P$ be the surjective map given in Lemma 3.3. Since $P$ is projective, we can construct $v:P\to \underset{I}{⨁}R$ such that $uv$ is the identity map, $1$, on $P,$ by filling in the following diagram: $P$ $⊕IR$ $P$ $0$ $v$ $1$ Let $K=\mathrm{ker}u.$ Define $f:P\oplus K\to \underset{I}{⨁}R$ by $f\left(p,k\right)=v\left(p\right)$ and $g:\underset{I}{⨁}R\to P\oplus K$ by $g\left(r\right)=\left(u\left(r\right),w\left(r\right)\right)$ where $w:\underset{I}{⨁}R\to K$ is given by $w=1-vu.$ We need to check that $\mathrm{im}w\subseteq \mathrm{ker}u$ to make sure that $w$ and $g$ is well defined. Since $uv=1,$ $uw=u\left(1-vu\right)=u-uvu=u-u=0,$ and so $\mathrm{im}w\subseteq \mathrm{ker}u.$ $fg\left(r\right)=f\left(u\left(r\right),w\left(r\right)\right)=vu\left(r\right)+r-vu\left(r\right)=r$ and $gf\left(p,k\right)=g\left(v\left(p\right)+k\right)=\left(u\left(v\left(p\right)+k\right),w\left(v\left(p\right)+k\right)\right)=\left(uv\left(p\right)+u\left(k\right),v\left(p\right)-vuv\left(p\right)+k-vu\left(k\right)\right)=\left(p+0,0+k-v\left(0\right)\right)=\left(p,k\right)$ for and $k$ elements of and $K$ respectively. Thus, $f$ and $g$ are inverses and define an isomorphism between $P\oplus \mathrm{ker}u$ and $\underset{I}{⨁}R.$ Conversely, suppose that $P$ is such that $P\oplus Q=F$ for some $R-$module $Q$ and some free $R-$module $F.$ Suppose that $h\text{'}:P\to M\text{'}\text{'}$ and a surjective $f:M\to M\text{'}\text{'}$ are given. To show that $P$ is projective, we must show that there exists a $g\text{'}:P\to M$ such that $h\text{'}=fg\text{'},$ i.e. by filling in the following diagram: $P$ $M$ $M''$ $0$ $g'$ $h'$ $f$ Let $h:F\to M\text{'}\text{'}$ be given by $h\left(p,q\right)=h\text{'}\left(p\right)$ for $\left(p,q\right)\in F=P\oplus Q.$ Since every free module is projective, we know that $g:F\to M$ can be constructed so that $h=fg.$ $F=P⊕Q$ $M$ $M''$ $0$ $g$ $h$ Define $g\text{'}:P\to M$ by $g\text{'}\left(p\right)=g\left(p,0\right).$ Then $fg\text{'}\left(p\right)=h\left(p,0\right)=h\text{'}\left(p\right),$ and so $P$ is projective. $\square$

Using the results of these propositions the projective module of many rings can be identified.

Examples.

1. If $R=ℤ$ then every projective is free.
2. If $R$ is a field then every module is projective.
3. If $R={ℤ}_{6}$ then there are projectives which are not free.

## Projective Resolutions

Chain complexes of projective modules make projective resolutions. This construction will allow us to compute homology of projective resolutions. We now develop this motivation precisely.

Let $P=\left\{{P}_{q}\right\}$ be a chain complex of $R-$modules, then $P$ is a chain complex over $M$ if there is an $R-$map $\epsilon :{P}_{0}\to M$ that is surjective. The map is an augmentation and the chain complex ${P}^{#}=\left\{{P}_{q},M\right\}$ is an augmented chain complex over $M.$

A chain complex over $M$ is projective if each ${P}_{q}$ is projective.

A projective resolution of $M$ is a projective acyclic chain complex over $M.$

A definition has no motivation unless it has some application. The first step towards motivating the definition of projective resolutions is the following proposition.

Projective resolutions exist.

 Proof. A projective resolution $P$ of $M$ exists if and only if there exists an exact sequence of projective modules ending with $M.$ We construct a sequence of free modules $\left\{{F}_{i}\right\}$ and maps ${p}_{i}:{F}_{i}\to {F}_{i-1}$ such that ${p}_{0}:{F}_{0}\to M$ is surjective and $\mathrm{ker}{p}_{i-1}=\mathrm{im}{p}_{i}$ for all $i.$ This sequence forms an exact sequence of projective modules ending with $M.$ The ${F}_{i}$ and the ${p}_{i}$ are constructed by induction. Lemma 3.3 allows us to find ${F}_{0}$ and ${p}_{0}$ such that ${p}_{0}:{F}_{0}\to M$ is surjective and ${F}_{0}$ is free. For the inductive step assume that the ${F}_{i}$ and the ${p}_{i}:{F}_{i}\to {F}_{i-1}$ have been defined such that $\mathrm{ker}{p}_{i-1}=\mathrm{im}{p}_{i}$ for all $i\le n.$ By Lemma 3.3 there exists a free module ${F}_{n+1}$ and a surjective map $u:{F}_{n+1}\to \mathrm{ker}{p}_{n}.$ Let $i:\mathrm{ker}{p}_{n}\to {F}_{n}$ be the imbedding of $\mathrm{ker}{p}_{n}$ in ${F}_{n}.$ $i$ is the restriction of the identity map on ${F}_{n}$ to $\mathrm{ker}{p}_{n}.$ So, the image of $i$ is $\mathrm{ker}{p}_{n}.$ Define ${p}_{n+1}:{F}_{n+1}\to {F}_{n}$ by ${p}_{n+1}=iu.$ Then $\mathrm{im}{p}_{n+1}=\mathrm{ker}{p}_{n}$ and the induction is complete. $\square$

The following results will be used to show that Tor is well defined. Because of this, they are especially significant.

Let $P$ and $P\text{'}$ be chain complexes over $M$ and $N$ respectively. Let $f:M\to N$ be an $R-$map. Then a chain map $f:P\to P\text{'}$ is a chain map over $f$ if the following diagram commutes (i.e. if $f:{P}^{#}\to {P\text{'}}^{#}$ is a chain map).

Let $f:M\to N$ be an $R-$map. let $P$ and $P\text{'}$ be chain complexes over $M$ and $N$ respectively. Assume that $P$ is projective and that $P\text{'}$ is acyclic.

1. There is a chain map $\left\{{f}_{i}\right\}$ over $f.$
2. Any two chain maps over $f$ are chain homotopic.

 Proof. (i) We construct $\left\{{f}_{i}\right\}$ by induction. Since $ϵ:{P}_{0}\text{'}\to N$ is surjective and ${P}_{0}\text{'}$ is projective we can construct ${f}_{0}:{P}_{0}\to {P}_{0}\text{'}$ such that ${f}_{0}ϵ\text{'}=ϵf$ by filling in the following diagram: $P$ $M$ $P0$ $N$ $0$ $f0$ $ϵ$ $f$ $ϵ'$ For the inductive step, assume that we have constructed the ${f}_{i}$s for $i\le n$ and that ${\partial }_{i}\text{'}{f}_{i}={f}_{i-1}{\partial }_{i}$ for all $i.$ Notice that since $P\text{'}$ is acyclic $\mathrm{im}{\partial }_{n+1}\text{'}=\mathrm{ker}{\partial }_{n}\text{'},$ and since ${\partial }_{n}\text{'}{f}_{n}{\partial }_{n+1}={f}_{n-1}{\partial }_{n}{\partial }_{n+1}={f}_{n-1}0=0,$ $\mathrm{im}{f}_{n}{\partial }_{n+1}\subset \mathrm{ker}{\partial }_{n}\text{'}=\mathrm{im}{\partial }_{n+1}\text{'}.$ Thus, since ${P}_{n+1}$ is projective, ${f}_{n+1}$ can be constructed by filling in $Pn+1$ $Pn$ $P'n+1$ $im ∂'n+1$ $0$ $fn+1$ $∂n+1$ $fn$ $∂'n+1$ so that ${\partial \text{'}}_{n+1}{f}_{n+1}={f}_{n}{\partial }_{n+1}.$ Thus there is a chain map over $f.$ (ii) We need to show that there exists a ${D}_{q}:{P}_{q}\to {P\text{'}}_{q+1}$ for all $q$ such that ${\partial \text{'}}_{q+1}{D}_{q}+{D}_{q+1}{\partial }_{q}={f}_{q}-{g}_{q}.$ Since ${f}_{-1}={g}_{-1}=f,$ we can define ${D}_{-1}=0.$ We define the rest of the ${D}_{q}$ by induction. Note that $ϵ\text{'}\left({f}_{0}-{g}_{0}\right)=ϵ\text{'}{f}_{0}-ϵ\text{'}{g}_{0}=f-f=0.$ Since $P\text{'}$ is acyclic $\mathrm{im}{\partial \text{'}}_{1}=\mathrm{ker}ϵ\text{'}$ and $\mathrm{im}\left({f}_{0}-{g}_{0}\right)\subseteq \mathrm{im}{\partial \text{'}}_{1}.$ Since ${P}_{0}$ is projective and ${\partial \text{'}}_{1}{D}_{0}={\partial \text{'}}_{1}{D}_{0}+{D}_{-1}ϵ={f}_{0}-{g}_{0},$ ${D}_{0}$ can be constructed by filling in the following diagram: $P0$ $P'1$ $im∂'1⊆P0$ $D0$ $f0-g0$ For the inductive step assume that ${D}_{q}$ has been constructed appropriately for all $q\le n.$ Notice that $∂'n+1(fn+1-gn+1) = ∂n+1'fn+1 - ∂n+1'gn+1 = fn∂n+1 - gn∂n+1 = (fn-gn)∂n+1 = (∂n+1'Dn + Dn-1∂n)∂n+1 = ∂n+1'Dn∂n+1+0.$ So ${\partial }_{n+1}^{\text{'}}\left({f}_{n+1}-{g}_{n+1}-{D}_{n}{\partial }_{n+1}\right)=0.$ Thus $\mathrm{im}\left({f}_{n+1}-{g}_{n+1}-{D}_{n}{\partial }_{n+1}\right)\subseteq \mathrm{ker}{\partial }_{n+1}^{\text{'}}.$ Since $P\text{'}$ is acyclic, $\mathrm{ker}{\partial }_{n+1}^{\text{'}}=\mathrm{im}{\partial }_{n+1}^{\text{'}}$ and $\mathrm{im}\left({f}_{n+1}-{g}_{n+1}-{D}_{n}{\partial }_{n+1}\right)\subseteq \mathrm{im}{\partial }_{n+2}^{\text{'}}.$ Since ${P}_{n+1}$ is projective, the following diagram can be filled in such that ${\partial }_{n+1}^{\text{'}}{D}_{n+1}={f}_{n+1}-{g}_{n+1}-{D}_{n}{\partial }_{n+1}.$ Thus ${\partial }_{n+2}^{\text{'}}{D}_{n+1}-{D}_{n}{\partial }_{n+1}={f}_{n+1}-{g}_{n+1}$ and the induction is complete. $Pn+1$ $Pn+2'$ $im ∂n+2'⊆Pn+2$ $Dn+1$ $fn+1-gn+1-Dn∂n+1$ So $\left\{{f}_{i}\right\}$ and $\left\{{g}_{i}\right\}$ are homotopic. $\square$

Any two projective resolutions over $M$ are chain equivalent.

 Proof. Let $P=\left\{{P}_{i}\right\}$ and ${P}^{\text{'}}=\left\{{P}_{i}^{\text{'}}\right\}$ be projective resolutions over $M.$ Since the identity map from $M$ to $M$ is an $R-$map, by the above theorem we can construct a chain map $\left\{{f}_{i}\right\}$ from $P$ to ${P}^{\text{'}}$ over the identity map. Similarly we can construct a chain map $\left\{{g}_{i}\right\}$ from ${P}^{\text{'}}$ to $P$ over the identity. The composition of these two maps is a chain map $\left\{{g}_{i}{f}_{i}\right\}$ from $P$ to $P$ over the identity. The identity mapping $P$ to $P$ is also a chain map over the identity. By the above theorem $\left\{{g}_{i}{f}_{i}\right\}$ is chain homotopic to the identity map from $P$ to $P.$ Similarly $\left\{{f}_{i}{g}_{i}\right\}$ is chain homotopic to the identity map from ${P}^{\text{'}}$ to ${P}^{\text{'}}.$ $\left\{{f}_{i}\right\}$ and $\left\{{g}_{i}\right\}$ form a chain equivalence between $P$ and ${P}^{\text{'}}.$ $\square$

This result completes our introduction of projective resolutions. It is the primary step towards making a definition of Tor. Without it Tor would not be a well defined functor.

## Tor and Its Properties

We now define Tor. Tor is just the homology of a projective resolution tensored with an $R-$module. It is a functor of two $R-$modules: the module that the projective resolution is over, and the module that the projective resolution is tensored by. The results of the previous section show that Tor does not depend on the projective resolution used, but only on the module it is over. This makes Tor a well defined functor of two $R-$modules.

Let $M$ and $N$ be $\mathrm{R-}$modules. Let $P$ be a projective resolution of $M.$ $P=⋯ →0000Pq →0000Pq-1 →0000⋯ →0000P0 →0000M →00000.$ Define the chain complex $P\otimes N$ by $⋯→0000 Pq⊗N →0000 Pq-1⊗N →0000⋯→0000 P0⊗N →00000.$ The functor Tor is defined by ${\mathrm{Tor}}_{q}\left(M,N\right)={H}_{q}\left(P\otimes N\right).$

The following corollaries of Theorem 4.6 show that Tor is well defined and does not depend on $P.$

Let $f$ and $g$ be the chain equivalence between two projective resolutions $P$ and ${P}^{\text{'}}$ over $M$ as given in Theorem 4.6. Let $1$ be the identity map between $N$ and itself. Then $f⊗1: P'⊗N→ P⊗N and g⊗1: P⊗N→ P'⊗N$ are chain maps and $(f⊗1) (g⊗1)$ is homotopic to the identity between $P\otimes N$ and itself, and $(g⊗1) (f⊗1)$ is homotopic to the identity between ${P}^{\text{'}}\otimes N$ and itself.

${\left(f\otimes 1\right)}_{*}:{H}_{q}\left(P\otimes N\right)\to {H}_{q}\left({P}^{\text{'}}\otimes N\right)$ is an isomorphism.

This proves that Tor is a well defined functor. ${\mathrm{Tor}}_{1}\left(M,N\right)$ is usually written simply as $\mathrm{Tor}\left(M,N\right).$

Several properties of Tor are immediate. We derive some of these. Most of these results arise from special properties of the ring $R.$ The following proposition holds for all $R-$modules.

${\mathrm{Tor}}_{0}\left(M,N\right)$ is isomorphic to $M\otimes N.$

 Proof. If $P=\left\{{P}_{q}\right\}$ is a projective resolution then the sequence $⋯→0000 P2 →∂2 P1 →∂1 P0 →ϵ M→00000$ is exact. Tensoring by $N$ gives an exact sequence $P1⊗N →∂1⊗1 P0⊗N →0000 M⊗N →00000.$ Thus $M⊗N = (P0⊗N) im(∂1⊗1) = H0(P⊗N).$ $\square$

The following propositions and examples apply to the ring $ℤ.$

If $R=ℤ$ then ${\mathrm{Tor}}_{q}\left(M,N\right)=0$ for all $q\ge 2.$

 Proof. If $R=ℤ$ then every projective $R-$module is free. Thus, if $F$ is free the sequence $0→0000 kerj→0000 F→0000 M→00000$ is exact and forms a projective resolution $P$ over $M.$ We see that ${P}_{q}=0,$ and that ${H}_{q}\left(P\otimes N\right)=0$ for $q\ge 2.$ $\square$

If $R=ℤ$ and $m\in ℤ$ then

 Proof. In general if $F$ is free and $0→0000 E→0000 F→0000 M→00000$ is an exact sequence of modules then $\mathrm{Tor}\left(M,N\right)=\mathrm{ker}\left(i×1\right),$ where $1$ is the identity map on $N.$ If $P=0 →0000ℤ →×mℤ →0000ℤ/m →00000$ ($×m$ is the map given by multiplication by $m$) then $P\otimes N$ is $0→0000 N→×m N→00000.$ The proposition follows. $\square$

Summarizing, we have the following properties of Tor for $ℤ-$modules (abelian groups):

1. ${\mathrm{Tor}}_{0}\left(M,N\right)=M\otimes N.$
2. ${\mathrm{Tor}}_{q}\left(M,N\right)=0$ for $q\ge 2.$
3. $\mathrm{Tor}\left(ℤ,N\right)=0.$
4. $\mathrm{Tor}\left(ℤ/m,N\right)=\mathrm{ker}\left(N\stackrel{×m}{\to }N\right)$ for $m\in ℤ.$
5. $\mathrm{Tor}\left(ℤ/m,ℤ/n\right)=ℤ/\mathrm{gcd}\left(m,n\right)$ for $m,n\in ℤ.$

If $R$ is a field, then ${\mathrm{Tor}}_{q}\left(M,N\right)=0$ for all $q>0.$

 Proof. 10 If $R$ is a field, an $R-$module is a vector space. Thus, every $R-$module has a basis and is free. So $0→0000 M→0000 M→00000$ is a projective resolution over $M.$ $\square$

These results completely describe the action of Tor on abelian groups and vector spaces.

$\mathrm{Tor}\left(M,N\right)$ has a very important property. It is commutative. We will prove this only in the case $R=ℤ,$ but it is true in general. The following two results are important tools for working with Tor. We will use them to show that Tor is commutative over abelian groups. Both of these results derive from the long exact sequence theorem of homology.

Suppose that $0\stackrel{\phantom{0000}}{\to }{N}^{\text{'}}\stackrel{\phantom{0000}}{\to }N\stackrel{\phantom{0000}}{\to }{N}^{\text{'}\text{'}}\stackrel{\phantom{0000}}{\to }0$ is an exact sequence of $R-$modules. Then there exists a long exact sequence $⋯→0000 Torq(M,N') →0000 Torq(M,N) →0000 Torq(M,N'') →0000 Torq-1(M,N') →0000⋯$ for every module $M.$

 Proof. The proposition follows from the long exact sequence theorem for homology provided that $0→0000 P⊗N' →0000 P⊗N →0000 P⊗N'' →00000$ is an exact sequence of chain complexes. Since $0→0000 N' →iN →0000 N'' →00000$ is exact, $Pq⊗N' →0000 Pq⊗N →0000 Pq⊗N'' →00000$ is exact. We show that since $i:{N}^{\text{'}}\to N$ is injective, so is ${i}^{\text{'}}:{P}_{q}\otimes {N}^{\text{'}}\to {P}_{q}\otimes N,$ where ${i}^{\text{'}}=1\otimes i$ and $1$ is the identity on ${P}_{q}.$ Let ${i}^{\text{'}\text{'}}:{P}_{q}\otimes {N}^{\text{'}}\to {P}_{q}\otimes N$ be given by ${i}^{\text{'}\text{'}}=1\otimes i,$ where $1$ is the identity on ${P}_{q}.$ Given a projective module $P,$ there exists a free module $F=\underset{k}{⨁}R$ such that $P\oplus {P}^{\text{'}}=F$ for some module ${P}^{\text{'}}.$ Consider the map $i'⊗i'': (Pq⊗N') ⊕ (Pq⊗N') → (Pq⊗N) ⊕ (Pq⊗N)$ given by ${i}^{\text{'}}\oplus {i}^{\text{'}\text{'}}.$ Now $\left({P}_{q}\otimes {N}^{\text{'}}\right)+\left({P}_{q}\otimes {N}^{\text{'}}\right)=F\otimes {N}^{\text{'}}=\underset{k}{⨁}{N}^{\text{'}},$ and similarly $\left({P}_{q}\otimes N\right)\oplus \left({P}_{q}\otimes N\right)=\underset{k}{⨁}N.$ So ${i}^{\text{'}}\oplus {i}^{\text{'}\text{'}}=\underset{k}{⨁}i.$ Since $i$ is injective ${i}^{\text{'}}\oplus {i}^{\text{'}\text{'}}$ is injective. Therefore ${i}^{\text{'}}$ is injective. Thus, the sequence $0→0000 Pq⊗N' →0000 Pq⊗N →0000 Pq⊗N'' →00000$ is exact and the proposition follows from the long exact sequence theorem. $\square$

If $0\stackrel{\phantom{0000}}{\to }{M}^{\text{'}}\stackrel{\phantom{0000}}{\to }M\stackrel{\phantom{0000}}{\to }{M}^{\text{'}\text{'}}\stackrel{\phantom{0000}}{\to }0$ is an exact sequence of $R-$modules, then there exists a long exact sequence $⋯→0000 Torq(M',N) →0000 Torq(M,N) →0000 Torq(M'',N) →0000 Torq-1(M',N) →0000⋯$ for every module $N.$

 Proof. We know that projective resolutions ${P}^{\text{'}}$ and ${P}^{\text{'}\text{'}}$ exist over ${M}^{\text{'}}$ and ${M}^{\text{'}\text{'}}$ respectively. By diagram chasing it can be shown that there is a surjective map from ${P}_{0}^{\text{'}}\oplus {P}_{0}^{\text{'}\text{'}}$ to $M,$ and thus that ${P}^{\text{'}}\oplus {P}^{\text{'}\text{'}}$ is a projective chain complex over $M.$ Since $0→0000 P'→0000 (P'⊕P'') →0000 P''→00000$ is a split short exact sequence of chain complexes, $0→0000 P'⊗N →0000 (P'⊗P'')⊗N →0000 P''⊗N →00000$ is a short exact sequence of chain complexes. The proposition follows from the long exact sequence theorem. $\square$

The following proposition shows that Tor is commutative if $R=ℤ.$

If $R=ℤ,$ then $\mathrm{Tor}\left(M,N\right)=\mathrm{Tor}\left(N,M\right).$

 Proof. First we show that $\mathrm{Tor}\left(M,F\right)=0$ for any free module $F=\underset{k}{⨁}ℤ.$ $0→0000 kerj→0000 F→0000 N→00000$ is a projective resolution. $0→0000 kerj⊗F →0000 F⊗F →00000$ is the same as $0→0000 ⨁kkerj →0000 ⨁kF →00000.$ Computing homology on $\underset{k}{⨁}\left(0\stackrel{\phantom{0000}}{\to }\mathrm{ker}j\stackrel{\phantom{0000}}{\to }F\stackrel{\phantom{0000}}{\to }0\right)$ gives $\mathrm{Tor}\left(M,F\right)=0.$ The exact sequence $0→0000 G=kerj→0000 F→0000 N→00000$ and Propositions 5.8 and 5.9 give the exact sequences $0→0000 Tor(M,G) →0000 Tor(M,F) →0000 Tor(M,N) →0000 M⊗G →0000 M⊗F →0000 M⊗N →00000$ and $0→0000 Tor(G,M) →0000 Tor(F,M) →0000 Tor(N,M) →0000 G⊗M →0000 F⊗M →0000 N⊗M →00000.$ These reduce to $0→0000 0→0000 Tor(M,N) →0000 M⊗G→0000 M⊗F→0000 M⊗N→00000,$ and $0→0000 0→0000 Tor(N,M) →0000 M⊗G→0000 M⊗F→0000 M⊗N→00000.$ Thus, $\mathrm{Tor}\left(M,N\right)=\mathrm{Tor}\left(N,M\right).$ $\square$

This completes our discussion of Tor. We have defined Tor and derived some of its properties. The projective resolutions used to define Tor and the results necessary to show that Tor is well defined were preceded by the relevant definitions and results of basic algebra and algebraic topology. As a composite this forms a complete and precise exposition of Tor.

## Remarks

Tor is only one of the functors used in algebraic topology. Tensor products, Hom, Ext, and Tor all have similar properties over fields and $ℤ-$modules. We have used tensor products extensively in this paper. The definition of Ext is very similar to the definition of Tor. Hom is used to define cohomology. The most powerful immediate application of these functors is in proving the universal coefficient theorems for homology and cohomology.

Homological algebra yields many tools that are useful in the study of algebraic topology. The applications of this algebra are fascinating and every mathematician should study them. I, however, will leave the future education of the reader in the hands of Munkres [Mu] and Spanier [Sp].

I am indebted to Professor F. Peterson for teaching me everything I know about algebraic topology. Most of the material in this paper was taken from his lectures for 18.905 Algebraic Topology, at MIT. I owe special thanks to all my professors at MIT for teaching me the art of mathematics, a task that required lots of patience on their part.

## References

[AM] Atiyah, M.F., and MacDonald, I.G., Introduction to Commutative Algebra, Addison-Wesley, 1969.

[Mu] Munkres, J.R., Elements of Algebraic Topology, Benjamin/Cummings, 1984.

[P] Peterson, F., Lectures for 18.905 Algebraic Topology, MIT, Fall 1986.

[Ps] Peterson, F., "The Start of Homological Algebra": Problem Set 2 for 18.905 Algebraic Topology, MIT, Fall 1986.

[Sp] Spanier, E., Algebraic Topology, McGraw-Hill, 1966; Springer-Verlag, 1982.

These notes are taken from

[R] Ram, A., An Introduction to Tor: Definitions and Properties, 18.905 Algebraic Topology, F. Peterson, March 1, 1987.