Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 1 / c

Rows in the following matrix index the source, columns the target, in the order $X,Y,Z$. Solving the arrow-square equation for a quiver representation morphism givesIn particular, the nonzero morphisms between distinct representations are $X\to Z$ and $Z\to Y$, each a one-dimensional family of scalar multiples of the vertexwise inclusion or projection. A map $Y\to Z$ is forced to vanish at the source by the identity arrow of $Z$; similarly a map $Z\to X$ is forced to vanish at the target. Maps between $X$ and $Y$ are zero. Each endomorphism ring is $k$.

The extension complex of quiver representations for the one-arrow quiver isIts cokernel is ${\mathrm{Ext}}_Q^1(U,V)$. Substitution of the three representations givesHere the first argument is the quotient endpoint of a short exact sequence. Thus the only possible nonsplit endpoint pair is subobject $X$, quotient $Y$. The sequenceis nonsplit, since $Z$ is indecomposable. More explicitly, all extensions with these endpoints have a middle arrow $k\stackrel{\lambda }{\to }k$: $\lambda =0$ gives the split short exact sequence, while each $\lambda \ne 0$ gives a middle representation isomorphic to $Z$. With endpoint identifications fixed the extension classes form $k$; up to endpoint automorphisms all nonzero classes give this same nonsplit sequence. There are no other nonsplit sequences with the listed endpoints, including equal endpoints.