Category and Hom, Tensor Functors

Category

Definition

A category $\mathcal{C}$ consists of

• a class of object;
• for each pair $(A,B)$, a set ${\mathrm{Hom}}_{\mathcal{C}}(A,B)$, whose elements are called morphisms from $A$ to $B$;
• for each triple of objects $(A,B,C)$ one has a map $\mathrm{Hom}(A,B)\times \mathrm{Hom}(B,C)\to \mathrm{Hom}(A,C),(f,g)\to g\circ f$.

It is assumed that the objects and morphisms satisfy the following conditions.

• If $(A,B)\ne (C,D)$, then $\mathrm{Hom}(A,B)$ and $\mathrm{Hom}(C,D)$ are disjoint.
• If $f\in \mathrm{Hom}(A,B)$ and $g\in \mathrm{Hom}(B,C)$, $h\in \mathrm{Hom}(C,D)$, then $(hg)f=h(gf)$.
• (Unit) For every object $A$, we have an element $1_A\in \mathrm{Hom}(A,A)$ such that $f{1}_A=f$ for any $f\in \mathrm{Hom}(A,B)$ and $1_{A}g=g$ for any $g\in \mathrm{Hom}(B,A)$.

Definition

A category $\mathcal{D}$ is called a subcategory of a category $\mathcal{C}$ if

• the objects of $\mathcal{D}$ is a subclass of the objects of $\mathcal{C}$
• for any objects $A,B$ of $\mathcal{D}$, one has that
  • ${\mathrm{Hom}}_{\mathcal{D}}(A,B)\subseteq {\mathrm{Hom}}_{\mathcal{C}}(A,B)$;
  • $1_A$ and the product of morphisms for $\mathcal{D}$ is the same as for $\mathcal{C}$.

The subcategory $\mathcal{D}$ is called full if ${\mathrm{Hom}}_{\mathcal{D}}(A,B)={\mathrm{Hom}}_{\mathcal{C}}(A,B)$.

Definition

Let $\mathcal{C}$ be a category. Define ${\mathcal{C}}^{\mathrm{op}}$ as

• objects of ${\mathcal{C}}^{\mathrm{op}}$ are the objects of $\mathcal{C}$
• ${\mathrm{Hom}}_{{\mathcal{C}}^{\mathrm{op}}}(A,B)={\mathrm{Hom}}_{{\mathcal{C}}^{\mathrm{op}}}(B,A)$

satisfying that

• if $f\in {\mathrm{Hom}}_{{\mathcal{C}}^{\mathrm{op}}}(A,B)$ and $g\in {\mathrm{Hom}}_{{\mathcal{C}}^{\mathrm{op}}}(B,C)$, then $gf$ in ${\mathcal{C}}^{\mathrm{op}}$ equals $fg$ in $\mathcal{C}$; and
• $1_{{\mathcal{C}}^{\mathrm{op}}}=1_{\mathcal{C}}$.

Definition

Let $\mathcal{C}$ and $\mathcal{D}$ be two categories. A covariant functor $F$ form $\mathcal{C}$ to $\mathcal{D}$ satisfies

• for any object $X\in \mathcal{C}$, $F(X)$ is an object of $\mathcal{D}$;

• for any $f\in {\mathrm{Hom}}_{\mathcal{C}}(X,Y)$, $F(f)\in {\mathrm{Hom}}_{\mathcal{D}}(F(X),F(Y))$ such that

  $$F(g\circ f)=F(g)\circ F(f),\text{mbox}foranyf\in {\mathrm{Hom}}_{\mathcal{C}}(X,Y),g\in {\mathrm{Hom}}_{\mathcal{C}}(Y,Z)\text{\hspace{1em}(*)}$$

• $F({\mathrm{id}}_X)={\mathrm{id}}_{F(X)}$.

If $(\ast )$ is modified to

$$F(g\circ f)=F(f)\circ F(g),\text{mbox}foranyf\in {\mathrm{Hom}}_{\mathcal{C}}(X,Y),g\in {\mathrm{Hom}}_{\mathcal{C}}(Y,Z)$$

then $F:\mathcal{C}\to \mathcal{D}$ is called a contravariant functor. In particular, a contravariant functor from $\mathcal{C}$ to $\mathcal{D}$ is a covariant functor from ${\mathcal{C}}^{\mathrm{op}}$ to $\mathcal{D}$.

Remark. In general, if $R$ is non-commutative, ${\mathrm{Hom}}_R(M,N)$ is not an $R$-module. Notice that

$$(r_1{r}_2)f(m)=f(r_1{r}_{2}m)=(r_{1}f)(r_{2}m)=(r_2{r}_{1}f)(m)$$

and $r_1{r}_2,r_2{r}_1$ are not always equal.

Hom Functor

In the category of $R$-modules, Hom functors can be defined as either covariant or contravariant, depending on whether the argument appears in the first or second slot.

• Given $R$-module $M$, define

  $$\begin{array}{rl}{\mathrm{Hom}}_R(M,\cdot ):R\text{mbox}-module& \to \text{mbox}Ab,\\ N& \mapsto {\mathrm{Hom}}_R(M,N),\\ f& \mapsto h_f:t\mapsto f\circ t.\end{array}$$

  We can verify that ${\mathrm{Hom}}_R(M,\cdot )$ is a covariant functor.

• Similarly, given $R$-module $M$, define

  $$\begin{array}{rl}{\mathrm{Hom}}_R(\cdot ,M):R\text{mbox}-module& \to \mathrm{Ab},\\ N& \mapsto \mathrm{Hom}(N,M),\\ f& \mapsto h_f:t\mapsto t\circ f\end{array}$$

  We can verify that ${\mathrm{Hom}}_R(\cdot ,M)$ is a contravariant functor.

Similar results hold in the setting of bimodules.

Definition

Let $R$ and $S$ be rings with identity. An abelian group $M$ is a left $R$- right $S$-bimodule in case $M$ is both a left $R$-module and a right $S$-module for which the two scalar multiplication jointly satisfy $r(xs)=(rx)s$ for any $r\in R,x\in M,s\in S$.

Remark. Here are some notations:

• ${}_{R}M$: $M$ is a left $R$-module
• $M_R$: $M$ is a right $R$-module
• ${}_R{M}_S$: $M$ is a left $R$- right $S$-bimodule.

Let $U:={}_R{U}_S$ be a bimodule. Then for any $R$-module ${}_{R}M$,

• ${\mathrm{Hom}}_R(U,M)$ is a left $S$-module.
  • For any $f\in {\mathrm{Hom}}_R(U,M)$, define $(sf)(u)=f(us)$.
  • ${\mathrm{Hom}}_R(U,\cdot ):R\text{mbox}-module\to S\text{mbox}-module$ is a covariant functor;
• ${\mathrm{Hom}}_R(M,U)$ is a right module-$S$.
  • For any $g\in {\mathrm{Hom}}_R(M,U)$, define $(gs)(m)=g(m)s$.
  • ${\mathrm{Hom}}_R(\cdot ,U):R\text{mbox}-module\to \text{mbox}module-S$ is a contravariant functor.

Tensor Functor

Suppose $M={}_S{M}_R$ and $N={}_{R}N$ are two modules, then $M{\otimes }_{R}N$ is a $S$-module. Similarly, suppose $M=M_R$ and $N={}_R{N}_T$ are two modules, then $M{\otimes }_{R}N$ is a module-$T$.

If ${}_S{M}_R$ and ${}_R{N}_T$ are bimodules, then $M{\otimes }_{R}N$ is a left $S$- right $T$-bimodule with $s(m\otimes n)=(sm)\otimes n$ and $(m\otimes n)t=m\otimes (nt)$.

Proposition

Given a bimodule ${}_S{U}_R$, define

$$U{\otimes }_R\cdot :R\text{mbox}-module\to S\text{mbox}-module,M\mapsto U{\otimes }_{R}M,f\mapsto {\mathrm{id}}_U\otimes f$$

and it is a covariant functor.

Similarly, define

$$\cdot {\otimes }_{R}R:\text{mbox}module-S\to \text{mbox}module-R,M\mapsto M{\otimes }_{R}U,f\mapsto f\otimes {\mathrm{id}}_U$$

and it is also a covariant functor.

Exact Sequence and Exact Functors

Definition

Let $F:R\text{mbox}-module\to S\text{mbox}-module$ be a covariant functor.

• We say $F$ is left exact if for any exact sequence $0\to N\stackrel{\phi }{\to }M\stackrel{\psi }{\to }P$, the sequence $0\to F(N)\stackrel{F(\phi )}{\to }F(M)\stackrel{F(\psi )}{\to }F(P)$ is exact.
• We say $F$ is right exact if for any exact sequence $N\stackrel{\phi }{\to }M\stackrel{\psi }{\to }P\to 0$, the sequence $F(N)\stackrel{F(\phi )}{\to }F(M)\stackrel{F(\psi )}{\to }F(P)\to 0$ is exact.

Definition

A contravariant $G:R\text{mbox}-module\to \mathrm{Ab}$ is called left (rep. right) exact if it is a left (rep. right) exact as a covariant functor $(R\text{mbox}-module{)}^{\mathrm{op}}\to \mathrm{Ab}$.

Definition

A functor is called exact if it is both left exact and right exact.

is exact.

Link to original

Theorem

Let ${}_S{U}_R$ be a bimodule. The functor $U{\otimes }_R\cdot :R\text{mbox}-module\to S\text{mbox}-module$ is right exact.

\begin{proof}

Remark that Theorem 7.4 is same as ^60fa98. \end{proof}