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Proposition Claim: Let $A$ be a finite-dimensional algebra over a field $k$. Suppose T,T' are simple $A$-modules and that there is a non-split extension $M$ of $T$ by T'. Then $T$ and T' are composition factors of the same projective indecomposable module.

Proof. We have a short exact sequence 0\to T'\xrightarrow{\iota_1} M\xrightarrow{\pi_1} T\to 0.

Let $P_T$ be the projective indecomposable corresponding to $T$, so $P_T/\mathrm{rad}(P_T)\cong T$, i.e. we have another short exact sequence $ 0\to \mathrm{rad}(P_T)\xrightarrow{\iota_2}P_T\xrightarrow{\pi_2}T\to 0. $

By the universal property of projective modules, using the surjective maps $\pi_1$ and $\pi_2$, there is a unique map $\varphi:P_T\to M$ satisfying $\pi_1\circ \varphi=\pi_2$.

Since the second sequence is exact, we know $\pi_2\circ\iota_2=0$ and so $\pi_2\circ\varphi\circ\iota_1=0$. Notice that \ker\pi_2=T'.

Hence by the universal property of $\ker\pi_2$, there exists a unique nonzero map \psi:\mathrm{rad}(P_T)\to T'. Since T' is simple, $\psi$ is surjective and by the first isomorphism theorem \mathrm{rad}(P_T)/\ker\psi\cong T'. Hence $P_T$ has a filtration $ P_T\supset \mathrm{rad}(P_T)\supset \ker\psi\supset 0 $ with composition factors $T$ and T' as required.

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    φ need not be unique for projective modules. I believe it is at least very restricted for projective covers though (only differ by automorphisms fi$x$ing the top factor).2011-09-02

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Suppose

0 ---> T' ---> M ---> T ---> 0  

is a non-split extension with $T$ and T' simple. Let $P\to T$ be the projective cover of $T$, whose kernel is $\mathrm{rad}(P)$. You should show that there is a commutative diagram with exact rows

0 ---> rad(P) ---> P ---> T ---> 0           |         |      |          |         |      |          V         V      V 0 --->   T'   ---> M ---> T ---> 0  

in which the arrow $T\to T$ is the identity and the map \mathrm{rad}(P)\to T' is not zero.

Can you conclude from this that T' is a composition factor in $P$?

For an example where your argument fails, consider the path algebra of the quiver $\overset1\bullet\to\overset2\bullet\to\overset3\bullet$ There is an extension of the simple $S_1$ by the simple $S_2$, but the radical of the projective cover $P_1$ of $S_1$ has length $2$, so it cannot be isomorphic to $S_2$ (depending on your conventions on modules, you might need to interchange $1$s and $3$s in what I wrote....)

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    Have updated my question to a complete proof.2011-09-02