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 ?

  • $\begingroup$ Could you please explain exactly how you came to know that the morphism k existed and made the diagram commute? The rest of the proof is as clear as crystal. $\endgroup$ Jul 26, 2018 at 21:35

1 Answer 1


Assume that $(a,b) \in \text{Ker }s,$ that is, $k(a)=p'(b)$.

Since $p$ is surjective, one can choose $e_0\in E$ such that $p(e_0)=a$. .Denote $b_0=h(e_0)$. From the commutativity of the RHS square, it follows that $$p'(b_0)=p'(h(e_0))=k(p(e_0))=k(a)=p'(b),$$ hence $b-b_0 \in \text{Ker }p' = \text{Im }i'$.

Thus, there is $m \in M$ such that $h(i(m))=i'(m)=b-b_0$ (note that here the commutativity of the LHS square was used).

Put $e:=e_0+i(m)$.

Then $$h(e)=h(e_0)+h(i(m))=b_0+(b-b_0)=b, \\ p(e)=p(e_0)+p(i(m))=p(e_0)+0=a.$$ Thus, $(a.b)\in \text{Im }r$.


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