Category Theory for Programmers Chapter 5: Products and Coproducts

Before starting, it can help to look at a few examples of posets.

Another thing that was probably a bit confusing but worth distinguishing because the chapter did not do too well at it was the definition of a product. I’ll state it here exactly:

Let $\mathscr{A}$ be a category and $X,Y \in \mathscr{A}$. A product of $X$ and $Y$ consists of an object $P$ and maps

with the property that $\forall A \in \mathscr{A}$ and $f_1,f_2$*

there exists a unique map $\bar{f}:A \to P$ such that the graph (if you drew the mappings) commutes. The maps $p_1$ and $p_2$ are called projections.

What was not super clear was that a product on a partially ordered set is unique and only on posets are you able to call it the product.

Also, in Category Theory for the Sciences, there was a slogan to help garner some intuition about coproducts:

Any time behavior is determined by cases, there is a coproduct involved.

  1. Let \(x,y \in \mathbf{C}\) be terminal objects. Because \(x\) is terminal, \(\exists! f:y \to x\) and because \(y\) is terminal, \(\exists! g:x \to y\). The composition \(f \circ g:x \to x\) is an arrow, but by definition, there is one and only one arrow from \(x\) to \(x\), and being an object in a category, it must have an identity, so \(f \circ g = \mathrm{Id}_x\). Similarly, \(g \circ f:y \to y\) and there can only be one morphism from \(y\) to \(y\) and so \(g \circ f = \mathrm{Id}_y\). Therefore, \(x \cong y\). \(\blacksquare\)

  2. Let \(c\) be a product of \(a\) and \(b\). This means we utilize the universal construction and get the relations:
    • \(c \to a\)
    • \(c \to b\)
    • for any other \(c’ \to a\) and \(c’ \to b\), we have a unique mapping \(c’ \to c\)

    So the question here is, given a poset, what is the product of objects? First step is to define what the relation of our poset is.

    Let \(a \to b\) if \(a\) is an ancestor of \(b\). Then our relations read as:

    • \(c\) is an ancestor of \(a\)
    • \(c\) is an ancestor of \(b\)
    • for any other ancestor \(c’\) of \(a\) and \(b\), we have that \(c’\) is an ancestor of \(c\). This makes \(c\) the most immediate ancestor of both \(a\) and \(b\) (and does not exclude \(c\) being \(a\) or \(b\)). However, this does not really work. If we look at the product of two siblings, we are saying the mother is an ancestor of the father and vice-versa.

    In the comments, a relationship of \(a\) is the boss of \(b\) was given by the author. In this case, we would have the product of two teammates to be their immediate boss, but this also assumes no person has more than one boss.

    However, for a real example that works, let \(a \to b\) be given by \(a,b \in \mathbf{Set}\) and \(a \subseteq b\). Then we have that the product of \(a\) and \(b\) is a subset \(c\) such that for any other subset \(c’\), \(c’ \subseteq c\). So the product is the largest subset of both.

    Another way to think about it is look looking at the composition of morphisms from the text. \(p’ = p \circ m\) which looks like \(c’ \to b = c’ \to c \to b\) and it seems to say, for any path, there’s an object that you can insert between that works for both \(a\) and \(b\).

  3. The coproduct of two elements in a poset would be flipping the relations around. In the case of subsets, the product of sets \(a\) and \(b\) would be the set \(c\) such that for any other set \(c’\) that contains both \(a\) and \(b\), we have that \(c \subseteq c’\), or that the product is the smallest set that contains both \(a\) and \(b\). This is the opposite of the product which is the largest set that is a subset of both \(a\) and \(b\). I guess you could think of this as the lim-sup of subsets of both \(a\) and \(b\). WHereas the coproduct would be lim-inf of sets that contain \(a\) and \(b\).

  4. Implement the equivalent of haskell Either.
    public class HelloWorld {
      public static class Either<L, R> {
        public boolean isLeft;
        public final L left;
        public final R right;
        private Either(L left, R right) {
          this.isLeft = left != null;
          this.left = left;
          this.right = right;
        public static <L, R> Either<L, R> left(L left) {
          return new Either<>(left, null);
        public static <L, R> Either<L, R> right(R right) {
          return new Either<>(null, right);
      private static void test(Either<?, ?> something) {
        if (something.isLeft) System.out.println("Given left value");
        else System.out.println("Given right value");
      public static void main(String[] args) {
        final Either<Integer, String> left = Either.left(37);
        final Either<Integer, String> right = Either.right("Hello, World");
  5. Showing Either is a better coproduct than int with the following injections:
    • int i(int n) { return n; }
    • int j(bool b) { return b ? 0 : 1; } So given the arrows int → int and bool → int, for Either to be better, we need arrows int → Either and bool → Either via some arrow m:Either → int
  6. Given the two injections above, i and j, we have “too many” injections in that we’re surjective but not injective. So it’s possible for overlap as the domain is larger than the codomain (ie, bool (2) + integers (2^64), integers (2^64)). So in this instance, one is not really any better than another.