Sorry for my bad english.
Call a morphism $f$ constant if $fr=fs$ for all $r,s$. The definition of co-constant is dual.
I begin to study category and i have one doubt. In some books, the definition of zero morphism is that morphism constant and coconstant, but if the category have a zero object, we can define the zero morfism to be the unique morphism that factors through the zero object. Now, i was trying to prove the equivalence of this to definitions assuming that te category have a zero object. One of the implications i did ok, my problem is the implication "constant + coconstant" imply "factors troough the zero object", for this side i just used that the morphism was constant. Here my argument:
Let $C$ a category with zero object $Z$. Let $f: X \rightarrow Y$ a morfism constant and coconstant. We know that exist morphisms $\phi_X : X \rightarrow Z $, $\phi_Y : Y \rightarrow Z $, $\psi_X : Z \rightarrow X $ and $\psi_Y : Z \rightarrow Y $.
Now, we can conclude that
$\phi_Y \circ f = \phi_X$ and $f \circ \psi_X = \psi_Y$ ,
$f \circ \psi_X \circ \phi_Y \circ f = \psi_Y \circ \phi_X$
but, because f is constant we know that
$f = f \circ \psi_X \circ \phi_Y \circ f$.
Why i didn't use the fact that f is coconstant?