# Category Group has unique identity morphism

Let $\mathcal C$ be a category which as finite products and a terminal element $\ast$. A monoid is a pair $(G,e,m)$, where $m: G \times G \rightarrow G$ and $e: \ast \rightarrow G$ are morphisms satisfying appropriate commutative diagrams (see page 20 http://www.jmilne.org/math/CourseNotes/AGS.pdf).

How can we show that $e$ is the unique morphism $\ast \rightarrow G$ which satisfies the given commutativity? Obviously this should parallel the proof that the identity of a monoid is unique (if $e_1, e_2$ are identities, then $e_1 = e_1e_2 = e_2$), but trying to work this out for the morphism $e$ has eluded me.

The proof is the same one-liner $e_1=e_1e_2=e_2$, but written down using category theory language. The relevant categorical concept here is that of "generalized elements", wherein you treat all morphism with some fixed domain $X$ as "elements of type $X$". The idea is that if you have a statement true for elements of type $Y$, a morphism $X\to Y$ gives you the same statement true for elements of type $X$ that factor through elements of type $Y$.

Within this framework of generalized elements, you can think of the identity morphism $G\overset{\mathrm{id}}\to G$ as a "generic" element of $G$, since any element of type $X$ factors through the identity morphism. At the other extreme, you can think of a group identity $*\overset{e}\to G$ as a "global element" since for any object $X$ you obtain a unique type $X$ element $X\to *\overset{e}\to G$ since $*$ is the terminal object.

The defining property of a monoid identity $*\overset{e_i}\to G$ is that both composites $G\overset{(e_i,\mathrm{id})}\to G\times G\overset{m}\to G$ and $G\overset{(\mathrm{id},e_i)}\to G\times G\overset{m}\to G$ are the identity morphism $G\overset{\mathrm{id}}\to G$ (where $(\cdot,\cdot)$ is the unique product morphism, and we identify $*\overset{e_i}\to G$ with the unique composite $G\to *\overset{e_i}\to G$). In terms of generic elements, these commutative diagrams say directly that multiplying the generic element by the group identity element gives the generic element back.

Since any element factors through the generic element, and since a global element can be considered an element of any type, If you precompose with a type $X$ element $X\overset{g}\to G$, $G\overset{(e_i,\mathrm{id})}\to G\times G\overset{m}\to G$ and $G\overset{(\mathrm{id},e_i)}\to G\times G\overset{m}\to G$ become $X\overset{(e_i,g)}\to G\times G\overset{m}\to G$ and $X\overset{(g,e_i)}\to G\times G\overset{m}\to G$, while the resulting morphism $G\overset{\mathrm{id}}\to G$ becomes the type $X$ element $X\overset{g}\to G$. This says directly that multiplying a type $X$ element $g$ by a type $X$ group identity element $X\to *\overset{e_i}\to G$ results in that same type $X$ element.

In particular, the identity has to be unique as we recover $e_1=e_1e_2=e_2$.

• Thanks for your answer, but I don't get it. Why should we have $m \circ (g,e_i) = g$? – D_S Apr 1 '15 at 23:11
• A (two-sided) identity $e$ in a monoid $G$ is by definition an element so that the product with any other element $g$ is that same element, i.e. so that $e\cdot g=g$ and $g\cdot e=g$. In terms of the multiplication map $m\colon G\times G\to G$, this says exactly that $m\circ (e,g)=g$ and $m\circ(g,e)=g$. – Vladimir Sotirov Apr 2 '15 at 2:04
• I still don't see how it says this. How does what you claim follow from the definition of $e$? – D_S Apr 2 '15 at 5:10
• I have edited a bit to clarify. – Vladimir Sotirov Apr 2 '15 at 17:54

We can deduce this using the yoga of representable functors (which is really just a way of working with generalised elements).

Suppose $M$ is a monoid object in a category. It is easy to see that $e : {*} \to M$ and $m : M \times M \to M$ give each hom-set $\mathrm{Hom}(T, M)$ the structure of a monoid in the classical sense. In particular, $\mathrm{Hom}({*}, M)$ is a monoid, with unit element $e$. Now, suppose $e' : {*} \to M$ is also a unit element for $M$. Then $e'$ would be a unit element for $\mathrm{Hom}({*}, M)$. So we have $e = e e' = e'$ in $\mathrm{Hom}({*}, M)$, as required.