# 3.Mono-, Epi- and Isomorphisms

## 11.05.2020

### Contents/Index

1.Definition
2.Dual Category
@3.Mono-, Epi- and Isomorphisms
4.Initial and Terminal Objects
5.Products and Coproducts
6.Exponentation and Cartesian Closed Categories

### 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.