I’m currently reading through Ralf Schiffler’s Quiver Representations, but I’m having trouble with the proof of the theorem below. In the following, $Q$ denotes a finite, acyclic quiver with set of vertices $Q_0$, and for every $i \in Q_0$ we denote by $P(i)$ the corresponding indecomposable projective representation over the ground field $k$.
Theorem 2.24. Subrepresentations of projective representations in $\operatorname{rep} Q$ are again projective.
Proof. Suppose that $P$ is a projective representation with dimension vector $(d_i)_{i \in Q_0}$. We will prove the theorem by induction on $d = \sum_{i \in Q_0} d_i$, the dimension of $P$.
If $d = 1$, then $P$ is simple and there is nothing to prove. So suppose that $d > 1$. Let $M$ be a subrepresentation of $P$ and let $u \colon M \to P$ be the inclusion morphism. By Corollary 2.21, we have $P \cong P(i_1) \oplus \dotsb \oplus P(i_t)$ for some vertices $i_1, \dotsc, i_t$, and thus the inclusion $u$ is of the form $$ u = \begin{bmatrix} u_1 \\ \vdots \\ u_t \end{bmatrix} $$ with $\operatorname{im} u_j \subseteq P(i_j)$. It follows that $M \cong \operatorname{im} u_1 \oplus \dotsb \oplus \operatorname{im} u_t$, and by Propositition 2.7, it sufficies to show that $\operatorname{im} u_j$ is projective for each $j$. This is obvious in the case where $\operatorname{im} u_j = P(i_j)$, so let us suppose that $\operatorname{im} u_j$ is a proper subrepresentation of $P(i_j)$. Then $\operatorname{im} u_j$ is a subrepresentation of $\operatorname{rad} P(i_j)$, and $\operatorname{rad} P(i_j)$ is projective by Lemma 2.23. Moreover, the dimension of $\operatorname{rad} P(i_j)$ is stricly smaller than $d$, and, by induction, we conclude that $\operatorname{im} u_j$ is projective, which completes the proof.
I don’t understand why $M \cong \operatorname{im} u_1 \oplus \dotsb \oplus \operatorname{im} u_t$, and suspect that this might be wrong:
Consider the quiver $Q$ consisting of a single vertex, labeled $1$, and no edges. Then representations of $Q$ over $k$ are just $k$-vector spaces, and we have that $P(1) = k$. Then $P := k^2 = P(1) \oplus P(1)$ is a projective representation of $Q$, and the diagonal $M = \{(x,x) \mid x \in k\}$ is a subrepresentation of $P$. With respect to the decomposition $P = P(1) \oplus P(1)$ the inclusion $u \colon M \to P$ takes the form $$ u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} $$ where $u_i \colon M \to P(1) = k$ is the projection onto the $i$-th component. But then $$ M \cong k \ncong k^2 = \operatorname{im} u_1 \oplus \operatorname{im} u_2. $$
Question: (Why) do we have that $M \cong \operatorname{im} u_1 \oplus \dotsb \oplus \operatorname{im} u_t$?