Link Search Menu Expand Document

Caution

We are suppressing set-theoretic hazards for categories here. For a concise bit on universes and some motivation see this post on Stack Overflow.

Also, I am using the convention that statements with lower case letters have implicit universal quantifiers on them as needed by the context.

Definition of categories

A category $\mathcal C$ is

  • a set of objects $\operatorname{Ob} \mathcal C$
  • for any two $X, Y \in \operatorname{Ob} \mathcal C$, written also as $X, Y \in \mathcal C$, a set \(\operatorname{Hom}_{\mathcal C}(X,Y)\) where elements are written $f: X \to Y$,
  • and for any $X, Y, Z \in \mathcal C$, a composition \(\circ: \operatorname{Hom}_{\mathcal C}(X,Y) \times \operatorname{Hom}_{\mathcal C}(Y,Z) \to \operatorname{Hom}_{\mathcal C}(X,Z)\) satisfying
    • associativity $f \circ (g \circ h) = (f \circ g) \circ h$ and
    • unitality: for each $X \in \mathcal C$ there exists $1_X : X \to X$ so that for any $f: X \to Y$ and $g : Y \to X$ we \(f \circ 1_X = f \ , \ 1_X \circ g = g\)

Examples of categories

  • Aside from the empty category, the simplest category is one where \(\operatorname{Ob} \mathcal C = \lbrace \ast \rbrace\) and \(\operatorname{Hom}(\ast,\ast) = \lbrace 1_\ast \rbrace\) Composition is trivial.

  • A variant on the previous example is where we take $(N,\cdot)$ to be a monoid still take a single object but now set \(\operatorname{Hom}(\ast,\ast) = N\) where composition is the monoid operation. In some sense, categories are monoid-oids.

  • A category useful in algebraic topology has objects $\mathbb{N}$ with \(\operatorname{Hom}(n,m) = \begin{cases} \empty & n > m \\ f: [n] \to [m] & f \text{ non-decreasing } \end{cases}\)

  • One closer to linear algebra also has objects $\mathbb{N}$ but has \(\operatorname{Hom}(n,m) = \operatorname{Hom}_k (k^n, k^m)\) where $k$ is a field. Here the morphism spaces are matrices.

  • Given a category $\mathcal C$, we can make another, the opposite category, $\mathcal C^{op}$ whose objects are the same but with \(\operatorname{Hom}_{\mathcal C^{op}}(X,Y) = \operatorname{Hom}_{\mathcal C}(Y,X)\)

Reminder on modules

Let $R$ be a (unital) ring. Recall a left $R$-module $M$ is the data of an abelian group $(M,+)$ and an action map $\cdot : R \times M \to M$ so that the action is linear

  • $r \cdot (m_1 + m_2) = r \cdot m_1 + r \cdot m_2$
  • $(r_1 + r_2) \cdot m = r_1 \cdot m + r_2 \cdot m$, ie

the action satisfies

  • $(r_1r_2) \cdot m = r_1 \cdot (r_2 \cdot m)$ and
  • $1 \cdot m = m$

which expresses the fact that map $R \to \operatorname{End}(M)$ is a monoid homomorphism. We will often suppress the $\cdot$ notation for the action itself and just concatenate the symbols.

Given a two $R$-modules $M_1,M_2$, a $R$-module morphism $f : M_1 \to M_2$ is an abelian groups homomorphism satisfying

  • $f(r \cdot m) = r \cdot f(m)$.

Examples

  • Calling back to linear algebra, we can take $R = k$ a field and $M = k^n$ with scalar multiplication.

  • More important for our course is the following. Let $f : R \to S$ be a homomorphism of rings and let $N$ be an $S$-module. Then we can equip the underlying abelian group $N$ with an $R$-module structure by declaring \(r \cdot n := f(r) n.\) The resulting $R$-module will be denoted by $f_\ast N$ and is called the pushforward of $N$. (The language will be clearer when we get to some geometry.)

  • As an example of pushforward, consider the inclusion $i: \mathbb{R} \to \mathbb{C}$ and the $\mathbb{R}$-vector space $i_\ast \mathbb{C}$. We have $i_\ast \mathbb{C} \cong \mathbb{R}^2$ since $\dim_{\mathbb{R}} \mathbb{C} = 2$.

  • Any ideal $I \leq R$ is an $R$-module as is any quotient $R/I$.

Let us look at the set $\operatorname{Hom}_R(R,R)$. We have an evaluation map \(ev_1 : \operatorname{Hom}_R(R,R) \to R \\ f \mapsto f(1).\) The value at $1$ complemently determines $f$ since \(f(r) = f(r \cdot 1) = r f(1).\) We see that $f$ is given by right multiplication by $f(1)$. Since left multiplication and right multiplication commute, there is an inverse map \(R \to \operatorname{Hom}_R(R,R) \\ r \mapsto (r^\prime \mapsto r^\prime r).\)

In general, left multiplication is not an $R$-module homomorphism. For the function \(m_r (r^\prime) = r r^\prime\) saying that $m_r (r^\prime) = r^\prime m_r(1)$ is just saying that \(r^\prime r = r r^\prime\) which is only true for all $r^\prime$ if $r \in Z(R)$ the center of $R$.

Modules over the path algebra

We look at an extended example now. Let $Q$ be a finite direct graph, also known as a quiver. Let $R$ be a commutative ring. Then the path algebra $RQ$ is the free $R$-module on the set of paths in $Q$ with multiplication given by path concatenation.

For a particularly simple example, consider the following quiver

and use $Q$ to denote it also. Then we have four paths. The two arrows we will denote as $a$ and $b$ plus the length $0$ loops at the two nodes, $e_0$ and $e_1$. Thus \(RQ = Re_0 \oplus Ra \oplus Rb \oplus Re_1\) and the nonzero products are \(ae_0 = a = e_1a \ , \ be_0 = b = e_1b \ , \ e_0^2 = e_0 \ , \ e_1^2 = e_1.\)

Let $M$ be a $RQ$-module. Let us repackage the information present in the $RQ$-module structure in terms of more familiar data. Note that \(1 = e_0 + e_1\) in $RQ$. So we can write \(M = e_0 M \oplus e_1 M.\) For shorthand, set $M_i = e_i M$. Each $M_i$ is only an $R$-module. For each arrow $a$ in $Q$, action by $a$ gives a map \(\phi_a : M_{h(a)} \to M_{t(a)}\) which is $R$-linear. So we have an assignment \(M \mapsto \lbrace M_i, \phi_a : M_{h(a)} \to M_{t(a)} \mid i \in Q_0, a \in Q_1 \rbrace\) This process is reversible. Given $(M_i, \phi_a)$, we set \(M = \bigoplus_i M_i\) and let $a$ act by $\phi_a$ on $M$.

Given a morphism $f : M \to M^\prime$ of $RQ$-modules, since $f(e_i M) = e_i f(M)$ we see we have induced maps $f_i : M_i \to M_i^\prime$. Since $f(am) = af(m)$, we see that \(f_i \circ \phi_a = \phi_a \circ f_i\) for any arrow $a$. This process is also reversible.

Thus the data of $RQ$-modules and their morphisms is the same as collections of $R$-modules indexed by vertices with $R$-linear maps indexed by arrows of $Q$.

This is the first example of an equivalence of categories.