# Pushout of a diagram as a cokernel

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.

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

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 "$g_1+g_2$").