On page 124 of the book Elements of representation theory of associative algebras, volume 1, Example 3.10, I computed the modules in this example.

$$ S(3)=0\leftarrow 0 \rightarrow K \leftarrow 0, \\ P(2) = K\overset{1}{\leftarrow} K \overset{1}{\rightarrow} K \leftarrow 0,\\ P(4) = 0\overset{}{\leftarrow} 0 \overset{}{\rightarrow} K \overset{1}{\leftarrow} K,\\ P(2) \oplus P(4) = K\overset{1}{\leftarrow} K \overset{\left(\begin{matrix} 0 \\ 1 \end{matrix}\right)}{\rightarrow} K^{2} \overset{\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)}{\leftarrow} K. $$

Let $f_1: S(3) \to P(2)$ be the embedding and $f_2: S(3) \to P(4)$ be the embedding. Let $f=\left(\begin{matrix} f_1 \\ f_2 \end{matrix}\right): S(3) \to P(2) \oplus P(4)$. Then $f$ is injective. Is $f: S(3) \to P(2) \oplus P(4)$ the map in the sequence $$ 0 \to S(3) \overset{f}{\to} P(2) \oplus P(4) \to (P(2) \oplus P(4))/S(3) \to 0 \quad (1) $$ in Example 3.10?

I think that $$ (P(2) \oplus P(4))/(Im(S(3))) = (P(2) \oplus P(4))/(Im f) = K\overset{1}{\leftarrow} K \overset{0}{\rightarrow} 0 \overset{0}{\leftarrow} K, (2) \\ (P(2) \oplus P(4))/S(3) = K\overset{1}{\leftarrow} K \overset{1}{\rightarrow} K \overset{0}{\leftarrow} K. (3) $$ I am not sure about the maps in (3). Should $(P(2) \oplus P(4))/S(3)$ the sequence (1) in Example 3.10 be $(P(2) \oplus P(4))/(Im(S(3)))$? I think that the direct sum of $S(3)$ and (3) is $P(2) \oplus P(4)$ which contradicts the fact the (1) is non-split. Thank you very much.

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Their notation in the example is sloppy, but understandable if you keep the preceeding lemma in mind. Note that the end terms of an almost split sequence are both indecompoable modules, so both of your guess are incorrect. What they meant by $M=(P(2)\oplus P(4))/S(3)$ is the indecomposable module given by the non-split extension: $0 \to P_4 \to M \to P(2)/S(3) \to 0$.

To be slightly more precise yet intuitive, the quiver is: $1 \xleftarrow{\alpha} 2 \xrightarrow{\beta} 3 \xleftarrow{\gamma} 4$. The maps from $S(3) \to P(2)\oplus P(4)$ is $a \overline{e_3} \mapsto a(\beta+\gamma)$ for any $a\in kQ$. By using the philosophy as mentioned in Specific projective dimension of a module over bound quiver , the cokernel $M$ of the injection is by "tearing half of the $S(3)$ from $P(2)$ and half from $P(4)$, and then glue the remaining together", hence it is an indecompoable module with sincere support. You can verify this by first writing the basis of $M$, and then consider the action of $\beta$ and $\gamma$ on it.


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