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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.

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