I'm self-learning category theory at the moment and I'm having some trouble with a question regarding expressing a pushout as a particular case of a cokernel. I don't know how to do diagrams on here so bear with me.

Suppose we are working in a preadditive category.

We have arrows $g_1 : A\rightarrow B$ and $g_2 : A\rightarrow C$. The task is to show that the pushout of the diagram corresponding to these arrows is the cokernel of the map $g_1 + g_2 : A\rightarrow B + C$ along with the obvious arrows from $B$ and $C$ to it. Note that $+$ denotes the coproduct here. I'll use $\iota_j$ to denote the canonical injections.

I don't have any experience proving that the limit of two diagrams is the same but this is what I have so far:

Suppose $(Q, q)$ is the cokernel in question. Then I want to show that $(Q,q\circ \iota_1,q\circ\iota_2)$ is the pushout for $g_1$ and $g_2$.

Firstly, $q\circ\iota_1\circ g_1 = q\circ(g_1+g_2) = q\circ\iota_2\circ g_2$ so it commutes with the diagram so all that's left is to show that we have universality. These composites are also equal to $0_{AQ}$ but I'm not sure that's relevant.

So suppose we have an object $P$ and maps $p_1 : B\rightarrow P$ and $p_2 : C\rightarrow P$ such that $p_1\circ g_1 = p_2\circ g_2$. To show that $Q$ is universal with respect to this diagram I need to construct a unique map from $Q\rightarrow P$. Since $Q$ is the cokernel object for the map $g_1+g_2$ I want to use the universal property of the cokernel to get a unique map from $Q\rightarrow P$.

So to that end I tried to show that $\langle p_1, p_2\rangle \circ (g_1+g_2) = 0_{A,P}$, where $\langle p_1,p_2\rangle : B + C\rightarrow P$ is the map guaranteed by the definition of the coproduct.

But here's where I'm stuck, I don't know how to show that $\langle p_1, p_2\rangle \circ (g_1+g_2) = 0_{A,P}$.

Any tips would be appreciated.

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    $\begingroup$ What do you mean by the map $g_1+g_2:A\to B+C$? The universal property of the coproduct can only give you maps from the coproduct... $\endgroup$ – Arnaud D. Mar 12 '17 at 14:11
  • $\begingroup$ Could be that mathoverflow is better for this question. $\endgroup$ – mathreadler Mar 12 '17 at 14:33
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    $\begingroup$ @Amaud D. $g_1+g_2$ is the coproduct of morphisms, induced by the coproduct functor. $\endgroup$ – IAlreadyHaveAKey Mar 12 '17 at 14:40
  • $\begingroup$ @IAlreadyHaveAKey Then you have the wrong domain: the coproduct functor would give you a morphism $A+A\to B+C$. $\endgroup$ – Arnaud D. Mar 12 '17 at 14:45
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    $\begingroup$ And I'm pretty sure this is better here that at MO. $\endgroup$ – Arnaud D. Mar 12 '17 at 14:46

I assume that by "preadditive category" you mean, as wikipedia does, a category enriched over abelian groups. In order for your statement to make sense, you also need to assume that it has coproducts, which are therefore also biproducts. In this case, morphisms between biproducts and their composites can be expressed using a matrix calculus.

The more natural statement is that the pushout of $g_1:A\to B$ and $g_2:A\to C$ is the cokernel of the "column vector" map $(g_1,-g_2) : A \to B\oplus C$. Now a map $B\oplus C \to X$ is a "row vector" $[h,k]$ for $h:B\to X$ and $k:C\to X$, and their composite is $h g_1 + k (-g_2)$, or $h g_1 - k g_2$. To say that this is equal to zero, i.e. that $[h,k]$ factors through the cokernel of $(g_1,-g_2)$, is to say that $h g_1 = k g_2$, which is exactly the same condition for $h$ and $k$ to jointly factor through the pushout of $g_1$ and $g_2$. Thus, the cokernel coincides with the pushout.

To deduce the statement you asked about, note that $(-1) : C\to C$ is an isomorphism and $(-1) \circ g_2 = -g_2$. Therefore, the cokernel of $(g_1,-g_2)$ is isomorphic to the cokernel of $(g_1,g_2)$ (which I think is what your exercise means by "$g_1+g_2$").


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