give a counterexample of monoid

Consider the infinite Cartesian product

$$S = \prod_{i=1}^{\infty} \mathbb{Z} $$

The maps $f \colon S \rightarrow S$ form a monoid under composition, with identity being the identity map.

Let $a \colon S \rightarrow S$ be the map

$$ a ((z_1, z_2, z_3 \ldots)) = (z_2, z_3, z_4, \ldots)$$

Let $b$ be the map

$$ b ((z_1, z_2, z_3 \ldots)) = (0, z_1, z_2, z_3 \ldots)$$

and let $c$ be the map

$$ c ((z_1, z_2, z_3 \ldots)) = (1, z_1, z_2, z_3, \ldots)$$

Then $ab = ac = e$, but $b \neq c$.

Lets look at the endomorphisms of a particular vector space: namely let our vector space $V := \mathbb{R}^\infty$ so an element of $V$ looks like a countable (not necessarily convergent) sequence of real numbers. The set of linear maps $\phi: V \rightarrow V$ form a monoid under composition(prove it!). Let $R: V \rightarrow V$ be the right shift map, namely it takes $R: (x_0, x_1, \dots) \mapsto (0,x_0,x_1, \dots)$. Let $L: V \rightarrow V$ be the left shift map $L: (x_0, x_1, \dots) \mapsto (x_1, x_2, \dots)$, clearly $L \circ R = \textrm{id}_V = e$. Now define $R' : V \rightarrow V$ where $R' : (x_0, x_1, \dots) \mapsto (x_0, x_0, x_1, x_2, \dots)$. We also have $L \circ R' = \textrm{id}_V = e$, but these are different maps.

There are probably simpler examples, but this is pretty explicit so I thought it would be good to see.

Yes, there is a simpler example.

Let $G$ be the free monoid over $A =\{a,b,c\}$ satisfying the relations $ab = ac =e$.

Intuitively, $G$ is the set of all finite strings over $A$ that contain neither the subword $ab$ nor $ac$. Or, think of $b$ and $c$ as different but both of which cancel a single trailing $a$ when multiplied on the right.

If you are concerned that we must show $b \neq c$, then do it this way instead. Forget the part about relations and just define $G$ as strings over $A=${a,b,c}, that is: $G \subset A^*$, $G = \{x \in A^*:\text{neither }ab\text{ nor }ac\text{ is a subword of }x\}$ or $G = A^* - A^*abA^* - A^*acA^*$. Multiplication in $G$ is ordinary string catentation, except for two special cases: $xa \cdot by = xy$ and $xa \cdot cy = xy$ for $x, y \in G$. You can easily verify that $G$ with this multiplication is a monoid and meets the criteria, and there is no reason to worry that $b=c$.

In particular, the strings "$b$" and "$c$" are both in the free monoid $A^*$ and nothing in the subtraction of sets removes them, nor do any of the multiplication rules, so they are still in $G$. So therefore the monoid elements $b$ and $c$ in $G$ are distinct.

addendum -- This G is the "simplest" possible in the sense of being universal. That is, for any morphism $f$ and monoid $M$ satisfying the criteria $f(ab)=f(ac)=f(e)=e$, with $f:A^* \rightarrow M$, there is a morphism $g:G \rightarrow M$ with $g(x)=f(x) \text{ for all } x \in A$. $g$ simply acts properly on $A$ and then extends to $A^*$

That cannot be said of Barry's monoid of functions: $S \rightarrow S$ because it has extra structure, namely $ba=ca=e$ and others. So even if you use just the closure of $\{a,b,c,e\}$ under function composition (rather that all functions: $S \rightarrow S$), you'd have extra structure in S versus the free case, which prevents it from acting universal wrt G.

For anyone interested, here are a couple more examples, followed by some special cases for which this property is true.

Examples

1. Let $T$ be the monoid of functions on $\mathbb{N}$, under composition. Let $b(x) = 2x$, $c(x) = 2x+1$, and $a(x) = \lfloor x / 2 \rfloor$. We have $a(b(x)) = x$ and $a(c(x)) = x$ but $b \ne c$.

2. Let $X$ be the set of $\mathcal{C}^\infty$ functions on $[0,1]$, and let $L$ be the monoid of linear operators on $X$. (In fact, $L$ is a ring.) Let $A(f) = \frac{df}{dx}$, $B(f) = \int_0^x f(t) \; dt$, and $C(f) = f(0) + \int_0^x f(t) \; dt$. Again $A \circ B = A \circ C = \text{id}$, but $B \ne C$.

Special Cases

• Suppose $M$ is finite. Then $ab = ac = e$ implies $b = c$.

  Proof. $M$ naturally embeds in the monoid $T$ of functions on $M$ via $a \mapsto f_a$, where $f_a(x) = ax$. (See also.) $T$ is finite. If $ab = ac = e$ then $f_a$ is surjective, hence injective, hence its inverse is unique.

• Suppose $M$ is commutative. Then $ab = ac = e$ implies $b = c$.

  Proof. $b = abc = c$.