Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I have a short exact sequence $$0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0$$ of $G$-modules ($G$ any group) and I apply an exact additive contravariant functor $T$ to get $$0 \longrightarrow TC \longrightarrow TB \longrightarrow TA \longrightarrow 0,$$ then if I want to split my new exact sequence, by splitting the original sequence, do I have to split the original sequence from the left or the right? or does it not make a difference? I thought it might have to be from the left so that after applying $T$ I'd get something split from the right, but I dont know if this is right or even necessary.

Thank you

share|cite|improve this question
To be split exact is a property of a short sequence determined by the existence of two morphisms backwards and a list of equations which are preserved under any additive functor. Whether $T$ is covariant or contravariant doesn't matter and exactness is irrelevant. See Zhen Lin's answer here. – Martin Jan 26 '13 at 13:49
Also: Splitting lemma – Martin Jan 26 '13 at 14:01
up vote 5 down vote accepted

Recall the splitting lemma for additive categories:

For a short sequence $A \xrightarrow{f} B \xrightarrow{g} C$ the following are equivalent:

  1. The morphism $g$ is a cokernel of $f$ and $f$ has a left inverse $r$.
  2. The morphism $f$ is a kernel of $g$ and $g$ has a right inverse $s$.
  3. There are morphisms $A \xleftarrow{r} B \xleftarrow{s} C$ such that the identities $$\begin{align} rf &= 1_A & & rs = 0 \\ gf &= 0 & & gs = 1_C \end{align} \qquad \text{and} \qquad 1_B = fr + sg$$ hold.
  4. The sequence $A \xrightarrow{f} B \xrightarrow{g} C$ is isomorphic to the sequence $A \xrightarrow{i_A} A \oplus C \xrightarrow{p_C} C$.

None of the implications is particularly difficult to prove (see a bit below for a short outline).

The point is that the characterizations in 3. and 4. are easily verified to be preserved by applying an additive functor. Depending on which definition you use one or the other may be more adequate; it is not important whether $T$ is exact or not.

Suppose $T$ is contravariant and $f,g,r,s$ as in 3. Then we get $TA \xleftarrow{Tf} TB \xleftarrow{Tg} TC$ and $TA \xrightarrow{Tr} TB \xrightarrow{Ts} TC$ and applying $T$ to the equations in 3. one obtains after some rearrangement $$\begin{align} Ts Tg &= 1_{TC} & Ts Tr &= 0 \\ Tf Tg &= 0 & Tf Tr &= 1_{TA} \end{align} \qquad \text{and} \qquad 1_{TB} = Tg Ts + TrTf,$$ which is 3. in the target category. In particular, the left inverse $r$ of $f$ becomes the right inverse $Tr$ of $Tf$ and the right inverse $s$ of $g$ becomes the left inverse $Ts$ of $Tg$.

For example, 1. implies 3. as follows: We are given a left inverse $r$ of $f$, that is $rf = 1_A$. This implies that $f$ is monic and $r$ is epic. The equation $gf = 0$ follows from the assumption that $g$ is a cokernel of $f$. Put $h = 1_B - fr$. Then $hf = f - frf = 0$, so there is a unique $s \colon C \to B$ such that $h = sg$. Then the definition of $h$ with $h = sg$ gives $1_B = fr + sg$. To verify $rs = 0$ note that $rsg = rh = r - rfr = 0$ and $g$ is epic. To verify $gs=1_C$ note that $gsg= gh = g - gfr = g$ and $g$ is epic.

It's worth emphasizing again that $r$ determines $s$ uniquely via the equation $1_B - fr = sg$ and the cokernel property of $g$.

Similarly (or dually) 2. implies 3 and $s$ determines $r$ uniquely via the equation $1_B - sg = fr$ and the kernel property of $f$.

The identities in 3. imply that $\big[\smash{\begin{smallmatrix} r \\ g\end{smallmatrix}}\big] \colon B \to A \oplus C$ and $[f \; s] \colon A \oplus C \to B$ are inverse isomorphisms yielding an isomorphism of sequences as in 4. And the simple 4. implies 1. and 2. finishes the proof of the splitting lemma.

share|cite|improve this answer
Thank you, this helps. – Chris Birkbeck Jan 27 '13 at 17:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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