5
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – ErotemeObelus Jul 26 '18 at 21:35
3
$\begingroup$

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, $(a.b)\in \text{Im }r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.