Assume we have two exact sequences $$0 \to M \xrightarrow{\varphi} Q \xrightarrow{\psi} \mathcal{L} \to 0$$ and $$0 \to M \xrightarrow{\varphi'} Q' \xrightarrow{\psi'} \mathcal{L}' \to 0$$
where $M,Q,Q',\mathcal{L},\mathcal{L}'$ are $R$-modules and $Q,Q'$ are injective.
I want to show that $$Q \oplus \mathcal{L}' \cong Q' \oplus \mathcal{L}.$$
My reasoning:
We take note of the fact that if we have $R$-module homomorphisms $$\varphi:M \to Q$$ and $$\varphi':M \to Q'$$ so that we can form the pushout or fiber-sum
$$Y := Q \oplus Q'/\{(\varphi(m),-\varphi(m)\;|\; m \in M \}$$ such that the following diagram commutes
$\require{AMScd}$ \begin{CD} M @>\varphi'>> Q'\\ @V \varphi V V @VV \pi_2 V\\ Q @>>\pi_1> Y \end{CD}.
Now, we look at $$0 \to Q \xrightarrow{\pi_1} Y \xrightarrow{\psi' \circ p} \mathcal{L}' \to 0$$
where $$\pi_1(q) = (q,0)/\sim$$
Now, assume that $$\pi_1(q) = (q,0) = 0/\sim.$$ This implies that there exists an $m \in M$ such that $$(q,0) = (\varphi(m),-\varphi'(m))$$ but since $\varphi'$ is injective, $m$ must be $0$, and then $$(q,0) = (0,0).$$ Hence $\pi_1$ is injective.
Now, I want to show that $$\psi' \circ p:Y \to \mathcal{L}'$$ is a surjection, where $p$ is the projection $$p:Y \to Q'$$ together with the map $$\psi':Q' \to \mathcal{L}'.$$
Now, we know that $$\operatorname{Im}(\varphi') = \operatorname{ker}(\psi')$$ so that an element on the form $$Y \ni (\varphi(m),-\varphi'(m))/\sim \ \overset{\psi' \circ p}{\longmapsto} 0$$ while other elements $$Y \ni (q,q')/\sim \overset{\psi' \circ p}{\longmapsto} \psi'(q')$$ and since we know that $$\operatorname{Im}(\varphi') = \operatorname{ker}(\psi')$$ and we know that we can not have elements on the form $$(0,\varphi(m)) \in Y$$ since this forces $m = 0$ so that $$Y \ni (0,\varphi(m)) = (0,0) \in Y.$$
I believe this atleast argues for a well-defined map (although I am having some trouble showing it conclusively).
Then, since $\psi'$ is surjective, and $p$ is a projection, we get a surjection, and it´s exact at $Y$ (if I am correct).
Then since $Q$ is injective, the sequence splits, and we get that $$Y \cong Q \oplus \mathcal{L}'.$$
I believe one can similarly show, by looking at $$0 \to Q' \to Y \to \mathcal{L} \to 0$$ that $$Y \cong Q' \oplus \mathcal{L}$$ and the statemenent we wanted to prove follows.
Is my idea correct? If not, what is wrong? I am most unsure about that $$\psi' \circ p$$ is a well-defined map, and my argument for it.
Thank you in advance.
Edit: Actually, I think I need to define the projection as $$p:Y \to Q'\setminus \operatorname{Im}(\varphi')$$ otherwise, I don´t think it necessarily will be surjective.
Edit 2: Having thought some more, and reading up, I realize that since $Y$ is an $R$-module, and $\pi_2:Q' \to Y$ is an $R$-module homomorphism (I believe), we can just take the identity map $g:Q' \to Q'$ and then there exists a lift (by injectivity of $Q'$) $$F:Y \to Q'$$ such that $$F \circ \pi_2 = g.$$ Hence I believe I can change $$\psi' \circ p$$ to $$\psi' \circ F.$$
One way you can organize this so that it becomes readable is as follows.
First, suppose that you do the pushout for a pair of maps $A\to B$ and $A\to B'$, so you get a diagram $\require{AMScd}$ \begin{CD} A @>>> B' \\ @VVV @VVV\\ B @>>> X \end{CD} Show that if the map $A\to B'$ is injective, then its cokernel and the cokernel of $B\to X$ are isomorphic, and the latter map is also injective.
(You can prove with the benefit of not having all the context in your problem laying around. Here you will have to mess with elements, probably)
Given that, suppose you have two exact sequences $0\to M\to I\to N\to 0$ and $0\to M\to I'\to N'\to 0$. Then you can do the pushout of $M\to I$ and $M\to I'$, getting the square in the following diagram, which you can then complete by taking cokernels into the following diagram with exact columns and rows \begin{CD} {} @. 0 @. 0 \\ @. @VVV @VVV \\ 0 @>>> M @>>> I @>>> N @>>> 0 \\ @. @VVV @VVV @. \\ 0 @>>> I' @>>> X @>>> C @>>> 0 \\ @. @VVV @VVV \\ {} @. N' @. C' \\ @. @VVV @VVV \\ {} @. 0 @. 0 \end{CD} The observation above tells you that $N\cong C$ and $N'\cong C'$, and if you suppose that $I$ and $I'$ are injective then you know that the middle row and the middle column split and that therefore $$I\oplus N'\cong I\oplus C'\cong X\cong I'\cong C\cong I'\oplus N. $$
(This part of the argument can now be done completely in terms of maps and diagrams, without any need to even touch an element)