Identity morphism In a category these equations holds:
b = b * a
c = c * a
for some morphisms a,b,c;
a IS NOT an identity morphism.
Is this possible?
If the answer is yes can you show me some examples of such categories?
 A: I'm not sure which direction you're composing functions in, but if you take the category of sets, let b and c be constant functions, and let a be some permutation that fixes the values of b and c then you get an example working in either direction.
A: Use your equations to directly build a category. I see two examples:
Two objects
Your category has two objects, $X$ and $Y$. The morphisms of the category are presented by


*

*Generator $a : X \to X$

*Generators $b,c : X \to Y$

*Relation $b = ba$

*Relation $c=ca$


It's clear that this presents a category where $b \neq c$ and with


*

*$\hom(X,X) = \{ 1, a, a^2, a^3, \ldots \} $

*$\hom(X,Y) = \{ b, c \} $

*$\hom(Y,Y) = \{ 1 \}$


and the only nontrivial products are $ba^n = b$ and $c a^n = c$.
If we were lazy, we might add a relation $a^2 = a$ too, to simplify things.
One object
Your category has one object $X$. $\hom(X,X)$ is the monoid presented by generators $a,b,c$ and relations $b=ba$ and $c=ca$. It's not too hard to see that $b \neq c$ and $a \neq 1$.
Again, we could simplify; e.g. by adding the relation $xy=x$ whenever $x \neq 1$. Then the monoid is finite, with four distinct elements $\{1, a, b, c\}$.
(equivalently, this is monoid you get by adding a unit to the semigroup with three elements satisfying the identity $xy=x$)
