# Pushout,modules,short exact sequences

if we have this diagram in modules

$$\begin{array}{ccccccccc} & & 0 & & 0 & & & & \\ & & \downarrow & & \downarrow & & & & \\ 0 & \longrightarrow & K & \longrightarrow & P & \longrightarrow & L & \longrightarrow & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \longrightarrow & A & \longrightarrow & U & \longrightarrow & L & \longrightarrow & 0 \\ & & \downarrow & & \downarrow & & & & \\ & & F & & F & & & & \\ & & \downarrow & & \downarrow & & & & \\ & & 0 & & 0 & & & & \\ \end{array}$$

..00
0KPL0
0AUL0
..FF
..00

with KAF and KPL exact,and U is pushout,why also AUL and PUF are exact?

The pushout of a monic is monic in any abelian category, which accounts for the arrows $P\to U$ and $A\to U$. The arrows $U\to F$ and $U\to L$ are epic since the epics $A\to F$ and $P\to L$, respectively, factor through them. The maps out of $U$ kill the corresponding maps into $U$ by definition.
So all that remains to show is that the image of $P\to U$ contains the kernel of $U\to F$, and similarly for the other sequence. It's enough to show that every map $U\to M$ killing $P\to M$ admits a unique factorization $U\to F\to M$. Indeed, such a map induces a map $A\to M$ which kills $K\to A$ by commutativity of the pushout square, and so a unique factorization $A\to F \to M=A\to U\to F\to M$, inducing a factorization as desired by the pushout property of $U$. The argument for $U\to L$ is exactly symmetric.
• By the definition of the maps, actually. The only map we know to exist from $P\to F$ is the zero map, so the map $U\to F$ must be defined in terms of that and the given map $A\to F$. This determines a unique map by the pushout property since $0:P\to F$ and $A\to F$ agree on $K$. – Kevin Carlson Aug 30 '14 at 19:21