This is an exercise from Rotman, Introduction to homological algebra.
Given exact sequences of $R$-modules
\begin{array}{ccccccccc} 0 & \longrightarrow & M & \overset{i}{\longrightarrow} & E & \overset{p}{\longrightarrow} & Q & \longrightarrow & 0\\ 0 & \longrightarrow & M & \overset{i'}{\longrightarrow} & E' & \overset{p'}{\longrightarrow} & Q' & \longrightarrow & 0 \end{array}
where $E$ and $E'$ are injective, then there is an isomorphism $$Q \oplus E' \cong Q'\oplus E$$
What I have done:
I completed the diagram using diagram chasing and the injectivity of E'
\begin{array}{ccccccccc} 0 & \longrightarrow & M & \overset{i}{\longrightarrow} & E & \overset{p}{\longrightarrow} & Q & \longrightarrow & 0\\ & & id\downarrow & & h\downarrow & & k\downarrow\\ 0 & \longrightarrow & M & \overset{i'}{\longrightarrow} & E' & \overset{p'}{\longrightarrow} & Q' & \longrightarrow & 0 \end{array}
Then I tried to define an exact sequence
\begin{array}{ccccccccc} 0 & \longrightarrow & E & \overset{r}{\longrightarrow} & Q\oplus E' & \overset{s}{\longrightarrow} & Q' & \longrightarrow & 0\\ \end{array}
because in this case we could conclude $$Q\oplus E' \cong Q'\oplus E$$ due to the injectivity of $E$.
I defined $$r : E \to Q\oplus E'$$ $$e \mapsto (p(e),h(e))$$ $$s : Q\oplus E' \to Q'$$ $$(a,b) \mapsto k(a) - p'(b)$$
Then it's easy to see that $$\text{im}(r) \subseteq \ker(s)$$
But I can't show that $\ker(s) \subseteq \text{im}(r)$, what's wrong ?