# #2: Epic Morphisms

Given that the name of this site is Epic Math, I felt I should write about a topic where the word epic is a technical term! It is hardly surprising that such a term exists, as many otherwise non-mathematical words such as almost, simple, open, connected, regular, normal, field, ringonto, map, twin, lucky, and even sexy have technical definitions.

Category Theory

The term epic is found in category theory, which is an extremely abstract branch of mathematics that formally deals with many other fields of math. It is so strange that some mathematicians have labeled it “abstract nonsense.”

In category theory the main subject of study is the category, which consists of three parts:

• A class of objects,
• A class of morphisms which go from one object to another object, and
• A binary operator called the composition function of morphisms, satisfying associativity and existence of identity.

Do not worry if you do not know what objects or morphisms are—they are very abstractly defined ideas. Here are several examples to make the theory more concrete:

Example 1: Let us write three objects $A, B, C$. We can also write morphisms $f: A \to B$ and  $g: B \to C$. Since categories have a composition function, then there is also a morphism $h: A \to C$ such that $h = g \circ f$.

Example 2: Using the notation from the above example, let $A, B, C$ all be $\mathbb{R}$, the set of real numbers. Let the morphisms $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be functions from $\mathbb{R}$ to $\mathbb{R}$, which should be familiar to most readers. Then $h = g \circ f$ is just standard composition of functions, so $h(x) = (g \circ f)(x) = g(f(x))$. For instance, if $f(x) = 2x$ and $g(x) = x+1$, then $h(x) = g(f(x)) = g(2x) = 2x + 1$.

Example 3: Let $A = F^n$, the $n$-dimensional vector space over the field $F$. Let $B = F^m$ and $C = F^k$. Then let the morphisms $f : F^n \to F^m$ and $g : F^m \to F^k$ be linear transformations. Since they are linear transformations over finite vector spaces, the morphisms can be represented as matrices, i.e. let $[f]$ denote the $n \times m$ matrix of $f$, and let $[g]$ be the $m \times k$ matrix of $g$. Then the composition $h = g \circ f$ is given by the matrix $[h] = [g][f]$ with dimension $n \times k$ obtained from matrix multiplication. This is the correspondence between the composition of linear transformations and the multiplication of matrices seen in linear algebra.

Example 4: Let $A, B, C$ all be $\mathbb{R}^2$ (the plane), and let $f, g$ be rotations about the origin by some angle. Let the composition $g \circ f$ denote first applying $f$, then applying $g$. Then there is a rotation $h$ so that $h = g \circ f$. For instance, let $f$ be rotation by 90 degrees and $g$ be rotation by 180 degrees. Then $h$ is rotation by 270 degrees.

Example 5 (Abstract): Let $A, B, C$ all be the set of functions from $\mathbb{R}$ to $\mathbb{R}$, denote this set $\mathbb{R}^\mathbb{R}$. Then let $S, T$ be functions from $\mathbb{R}^\mathbb{R}$ to itself. Let composition be the standard composition of functions. Then given $S, T$ there is a function $P$ from $\mathbb{R}^\mathbb{R}$ to itself that is equal to $P = T \circ S$. For instance, let $S$ map all functions $f : x \mapsto y$ to $S(f) : x \mapsto (y + 1)$, i.e. $S$ increases every value of a function by 1. And let $T$ map all functions $f : x \mapsto y$ to $T(f) : x \mapsto 2y$, i.e. $T$ doubles every value of a function. Then $P = T \circ S$ maps all functions $f : x \mapsto y$ to $P(f) : x \mapsto (2y+1)$. For example, if $f(x) = \sin(x)$, then $(S(f))(x) = 2 \sin(x)$ and $(P(f))(x) = (T(S(f)))(x) = 2 \sin(x) + 1$.

Epic Morphisms

An epimorphism (also called an epic morphism) is a morphism $f : A \to B$ such that for all morphisms $g_1, g_2 : B \to C$, we have $g_1 \circ f = g_2 \circ f \implies g_1 = g_2$.

Here are a few basic useful properties of epimorphisms:

• Every isomorphism is an epimorphism.
• If $f$ and $g$ are epimorphisms, then $f \circ g$ is an epimorphism.
• If $f \circ g$ is an epimorphism, then $f$ is an epimorphism.
• The epic property is preserved under equivalence of categories.

Example 6: Let $A, B, C$ be $\mathbb{R}$, so our morphisms are functions. Then the function $f(x) = 2x$ is an epimorphism. Since $g_1 \circ f, g_2 \circ f$ are equal, then for all $x \in \mathbb{R}$$(g_1 \circ f)(x) = (g_2 \circ f)(x)$, i.e. that $g_1(2x) = g_2(2x)$. Substitute $y = 2x$, which is a surjective function. Then $g_1(y) = g_2(y)$ for all $y \in \mathbb{R}$, so $g_1 = g_2$ and thus $f$ is an epimorphism.

Example 7: Let $A, B, C$ be $\mathbb{R}$ as before, so our morphisms are functions. Then the function $f(x) = \sin(x)$ is not an epimorphism. For example, let $g_1(x) = 0$, and let $g_2(x)$ be the function that has value 0 in the interval $[-1, 1]$ and 1 everywhere else. Then clearly $g_1 \circ f = g_2 \circ f$ as both are equal to 0 for all $x \in \mathbb{R}$. However $g_1 \neq g_2$, so $f$ is not an epimorphism.

In fact, we can extend the process in Example 5 to show that every surjective function from a set to a set is an epimorphism, and every non-surjective function from a set to a set is not an epimorphism. Thus in the category of sets with functions, epimorphisms correspond to surjective functions. In general categories, however, epimorphisms do not always correspond to surjections. For example, in the category of rings with ring homomorphisms, the inclusion map $\mathbb{Z} \to \mathbb{Q}$ is an epimorphism but not a surjection.