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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?

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  • $\begingroup$ Perhaps this proves that if you have a zero object, constant is equivalent to coconstant? I don't know if that's true, but it doesn't seem absurd... $\endgroup$ – Ben Millwood Jan 12 '14 at 20:56
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    $\begingroup$ Please include your definition of constant and coconstant morphisms (not morpfisms by thr way). $\endgroup$ – Martin Brandenburg Jan 12 '14 at 21:09
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It's true that in a pointed category, that is one with a zero object, constant is equivalent to co-constant. One direction is below; the other is dual, since pointedness is self-dual.

Let $\mathcal{C}$ have a zero object $0$. Let $g:A\to B$ be constant, $h,k:B\to C$, and $i: 0\to A,f:A\to 0$. We want to show $hg=kg$. Since $g$ is constant, $g=g\circ 1_A=gif:A\to B$. Therefore $hg=hgif$ and $kg=kgif$. But since $0$ is final, $f$ is an epimorphism, whence we get $hgi=kgi$, so finally $kg=kgif=hgif=hg$ and indeed $f$ is co-constant.

This is related to the fact that in a pointed category, an object is a zero if and only if it's either initial or terminal.

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You haven't done anything wrong. To get $f = f \circ \psi_X \circ \phi_Y \circ f$ you could either use the definition of $f$ being constant applied to $\psi_X \circ \phi_Y \circ f$, or the definition of $f$ being coconstant applied to $f \circ \psi_X \circ \phi_Y$. From this you get that $f$ factors through the zero object, which you already proved implies constant+coconstant.

In summary, you've proved: $$f\ \text{factors through the zero object}$$ $$\Updownarrow$$ $$f\ \text{is constant}$$ $$\Updownarrow$$ $$f\ \text{is coconstant}$$ which is equivalent to proving what you wanted to prove.

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