Mono-, Epi- and Isomorphisms

Monomorphisms

An arrow $ f : A \rightarrow B $ in a category $ \mathcal{C} $ is a monomorphism (or monic) if for every pair of arrows, $ g,h : A \rightarrow B $ in $ \mathcal{C} $ we have that $$ f \circ g = f \circ h \Rightarrow g = h $$ In $ Set $ monotonically functions are monic. For example say we have the function from integers to integers: $$ f(x) = x^2 $$ this clearly isn't monic. Pick $$ g(x) = -x,h(x) = x $$ for which $$ f(g(2)) = f(h(2)) $$ but we do not have $$ g(2) = h(2) $$ On the other hand $$ f(x) = 2 \cdot x $$ is monic. Proof: let $ g,h $ be functions. In order to reach a contradiction assume $$ f \circ g = f \circ h \land g \neq h $$ now let $ x $ be such that $ h(x) \neq g(x) $. $ f $ is clearly injective. So we have that $$ x \neq y \Rightarrow f(x) \neq f(y) $$ And we have $$ f(g(x)) \neq f(h(x)) $$ contradicting our assumption.

In $ Set $ it is so that any monic arrow is injective. This goes for other structures as categories as well. Like $ Mon $,$ Grp $,$ Rng $ and so on

Epimorphism

An arrow $ f : A \rightarrow B $ is an epimorphism (or epic) if for any pair of arrow $ g,h : A \rightarrow B $, we have that $$ g \circ f = h \circ f \Rightarrow g = h $$ Again in $ Set $ these arrows are functions that are surjective. That is the functions for which $$ \forall y \in B\ \exists x \in A : f(x) = y $$

For the category of monoids we have some arrows that are epic, but not surjective. Observe the homomorphism $$ i : (\mathbb{N},+,0) \rightarrow (\mathbb{Z},+,0) $$ mapping each element of $ \mathbb{N} $ to the corresponding non-negative value of $ \mathbb{Z} $. $ i $ is not surjective. However given homomorhisms $ f,g : (\mathbb{Z},+,0) \rightarrow (M,\star,e) $ mapping to some monoid, we can show that $ i $ is epic. Let $ f \circ i = g \circ i $. Observe that for $ z \geq 0 $ we have that $ i(z) = z $. And hence $$ f(z) = f(i(z)) = g(i(z)) = g(z) $$ Now for $ z \lt 0 $ we have that $ -z = i(-z) $ and hence using the above that $ f(-z) = g(-z) $. And we get $$ f(z) = f(z) \star g(-z) \star g(z) = f(z) \star f(-z) \star g(z) = g(z) $$ Thus $ i $ is epic.

Isomorphism

An arrow $ f : A \rightarrow B $ is an Isomorphism if there is an arrow $ f^{-1} : B \rightarrow A $ such that $$ f \circ f^{-1} = id_B $$ and $$ f^{-1} \circ f = id_A $$ If any isomorphism exists between two objects $ A $ and $ B $ they are said to be isomorphic.