Additive functors preserve split exact sequences

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

Solution 1:

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.

Solution 2:

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.