How can I prove that additive functors preserve split exact sequences?

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    $\begingroup$ Prove that a split exact sequence $0 \to A \to B \to C \to 0$ is isomorphic to the obvious direct sum sequence $0 \to A \to A \oplus C \to C \to 0$. $\endgroup$ – t.b. Apr 18 '12 at 8:41
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    $\begingroup$ (Prove also that a functor is additive if and only if it preserves 0 and binary direct sums.) $\endgroup$ – Zhen Lin Apr 18 '12 at 8:42
  • $\begingroup$ @ZhenLin Please consider converting your comment (and the comment by t.b.) into a (hint only) answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Jun 18 '13 at 17:25

I am assuming we are dealing with a functor $F: R\mathrm{Mod} \to S\mathrm{Mod}$ where $R$ and $S$ are commutative rings, although the result may hold in more general settings that I am not sufficiently familiar with.

A split exact sequence $0 \to A \xrightarrow{i} B \xrightarrow{p} C \to 0$ can be characterized by 4 functions and 5 equations: \begin{align} i &: A \to B, \\ q &: B \to A, \\ j &: C \to B, \\ p &: B \to C, \\ q \circ i &= 1_A, \\ p \circ j &= 1_C, \\ p \circ i &= 0, \\ q \circ j &= 0, \\ i \circ q + j \circ p &= 1_B. \end{align} That is, the given sequence is split exact if and only if there are $j$ and $q$ so that $i, j, p, q$ satisfy the above equations. Now any functor preserves composition and identity, and additive functors also preserve addition and the 0 morphism, so the entire characterization of the split exact sequence is preserved, and hence its image is split exact.


Let $F:\mathcal{A}\to \mathcal{B}$ be an additive functor between abelian categories. A chain complex is split exact if and only if the identity map is null-homotopic (Weibel Exercise 1.4.3). Let $1_\mathcal{A}$ be the identity mapping from the chain complex $0\to A \stackrel{d}{\to} B \stackrel{d}{\to} C \to 0$ to itself, and $1_\mathcal{B}$ the identity mapping on $0\to F(A) \stackrel{F(d)}{\to} F(B) \stackrel{F(d)}{\to} F(C) \to 0$. Since the first sequence is split exact, we can find a chain contraction $s$ such that $1_\mathcal{A}=ds+sd$. Applying $F$ to both sides, we obtain $F(1_\mathcal{A})= 1_\mathcal{B} = F(d)F(s)+F(s)F(d)$. So the target sequence must also be split exact.


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