Monoid as a single object category I'm struggling with comprehending what monoids are in terms of category theory.
In examples they view integer numbers as a monoid. I think I get the set theoretic definition. We have a set and a associative binary operator (addition) and the neutral element (zero).
Then they are saying something like that - view the whole set as a single object and binary operator as bunch of morphisms for every element of the set.
Like add0 is an identity morphism. Which would really give us the same object i.e. the same set of all integer numbers. I think I understand this.
But let's view the morphism add1. After applying it to the our single object (the set of all integers) we would have a set {1,2,3…} not the {0,1,2,3…}. Aren't domain and codomain different in that case?
That's what's bothering me. Can someone clarify that to me?
Here is the text that gives me problems.
The text
 A: In addition to Henning Makholm's crisp and clear answer, you might find the opening six pages of my Notes on Category Theory helpful. They too give the example of a monoid as a category, but also give some other examples of categories where the arrows are not functions in any ordinary sense. Another important illustration is the case of a posets treated as a category.
In fact these examples suggest why we might well prefer to talk of 'arrows' rather than 'morphisms' (because the very term 'morphism' comes with baggage, and almost inevitably makes us think of a function -- but to repeat, arrows need not be functions).
A: No, that's not the way to view it. In order to view a monoid as a category, you have a single object $\mathsf{Andreas}$, and each element of the monoid is one morphism $\mathsf{Andreas}\to\mathsf{Andreas}$ in the category. The monoid operation is the composition in the category.
So for the integers, you don't have a morphism "add 1", but a morphism that is simply called $1$. And composition in the category works such that $1$ composed with $1$ is the morphism called $2$.
This is an example of a category where the morphisms are not functions.
A: You already know that a monoid $M$ is a set with a unit $e$ and a binary operation. More precisely, if $a,b,c\in M$ then $$a\circ b\in M$$ $$(a\circ b)\circ c=a\circ (b\circ c)$$ $$e\circ a=a\circ e=a$$
Now, take any category $C$ with one object, $c$. Since $C$ is a category, we need to say what the arrows are. That is, what the morphisms $c\rightarrow c$ are. There must be a unit $1_{c}:c\rightarrow c$, the arrows must be composeable and the composition must be associative. More precisely, if $f,g$ and $h$ are arrows, then $$f\circ g\in Morph(C)$$ $$(f\circ g)\circ h= f\circ (g\circ h)$$ $$1_{c}\circ f=f\circ 1_{c}=f$$Notice we are $\textit not$ talking about sets here. Just objects and arrows, in the abstract.
But now if we just observe that the operations on $M$ are $\textit exactly$ the same as the operations on $Morph(C)$, we may regard the category $C$ as the monoid $M$. This correspondence is reversible: given category $C=\left \{ c \right \}$ we obtain a monoid $M$ whose elements are the arrows of $C$. 
Thus the two descriptions are equivalent. 
All this works because the binary operations are the same for both structures. 
