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Let $X$ and $Y$ be indecomposable modules over a finite dimensional algebra and let $f \colon X \to Y$ be a non-zero morphism which is neither a monomorphism nor an epimorphism.

Suppose that it is possible to find a module $Z$ and morphisms $f'' \colon X \to Z$ and $g'' \colon Z \to Y$ such that $$ 0 \longrightarrow X \xrightarrow{(f' f'')} Im f \oplus Z \xrightarrow{(g' g'')} Y \longrightarrow 0 $$ is an exact sequence, where $f'$ is the corestriction of $f$ to its image and $g'$ is the inclusion.

How can I show that this sequence does not split?


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Let $Y=A\oplus B$ with $A\neq 0$ and let $f:X\rightarrow Y$ be any map with image $B$. Let $Z=X\oplus A$. Map $X$ to $Z$ as inclusion on the first factor and $Z$ to $Y$ as $-f$ on $X$ and he identity on $A$. Then your sequence is easily seen to split, which is why you can't show that it doesn't. – Steven Landsburg Jun 6 '14 at 0:52
$Y$ must be indecomposable. – Vasco Jun 6 '14 at 0:54
Are $X$ and $Y$ finite-dimensional? If so, then @WilberdvanderKallen's suggestion works. Or you can use the Krull-Schmidt Theorem: if the sequence splits, then $X\oplus Y\cong\operatorname{Im}(f)\oplus Z$, but $\operatorname{Im}(f)$ can't be isomorphic to either $X$ or $Y$ since it has smaller dimension than both. – Jeremy Rickard Jun 6 '14 at 10:56
@MarceloSilva: The Krull-Schmidt Theorem implies that every direct sum decomposition of $X\oplus Y$ into indecomposable modules has two summands isomorphic to $X$ and $Y$. But $\operatorname{Im}(f)\oplus Z$ can be refined to a direct sum decomposition into indecomposables. – Jeremy Rickard Jun 6 '14 at 12:41
@MarceloSilva: Every finite-dimensional module has a decomposition into indecomposables. If $\operatorname{Im}(f)=M_1\oplus\dots\oplus M_m$ and $Z=N_1\oplus\dots\oplus N_n$ are such decompositions, then $M_1\oplus\dots\oplus M_m\oplus N_1\oplus\dots\oplus N_n$ is a decomposition of $\operatorname{Im}(f)\oplus Z\cong X\oplus Y$. – Jeremy Rickard Jun 7 '14 at 7:04

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