give a counterexample of monoid If $G$ is a monoid, $e$ is its identity, if $ab=e$ and $ac=e$, can you give me a counterexample such that $b\neq c$? 
If not, please prove $b=c$.
Thanks a lot.

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.
