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I am given a short exact sequence of R-modules $$0\to A\to B\to C\to 0$$ such that it splits. The question is to show that the set of all splittings from $s:C\to B$ is in bijection with $\text{Hom}(C,A)$.

I have shown injection in one direction as follows. Since the sequence is split, we have $B\cong A\oplus C$ and thus $$\text{Hom}(C,B)=\text{Hom}(C,A\oplus C)\cong \text{Hom}(C,A)\oplus \text{Hom}(C,C)$$ and thus any section (which is essentially in $\text{Hom}(C,B)$) gives me a unique element in $\text{Hom}(C,A)$.

I am stuck on the other direction. Any help is appreciated!

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  • $\begingroup$ Sorry, I have edited the question. The question is for modules. $\endgroup$
    – HackR
    Aug 28, 2022 at 15:22
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    $\begingroup$ I think you could argue that the sections $s:C\to B$ correspond to elements $(f,g)$ of $$Hom(C,A)\oplus Hom(C,C)$$ such that $g=id_C$ (because for a section $s$ we want $(B\to C)\circ s=id_C$). $\endgroup$
    – Dave
    Aug 28, 2022 at 15:44
  • $\begingroup$ Actually, I guess it won't be $g=id_C$ in general, but it will be a specific map $g$ based on what the map $B\to C$ is in the exact sequence. The map $B\to C$ is really a map $A\oplus C\to C$ such that $A$ gets killed (since, exactness, $A$ is the kernel of $B\to C$). So restricting this to a map $C\to C$ we get an isomorphism, and we want $g$ to be the inverse of this map (I think!). $\endgroup$
    – Dave
    Aug 28, 2022 at 16:03

2 Answers 2

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Consider $0 \longrightarrow A \stackrel i \longrightarrow B \stackrel q \longrightarrow C \longrightarrow 0$. Let $j$ be a right inverse of $q$, and $S$ be the set of all right inverses of $q$. Note that for each $f$ in $\operatorname{Hom}(C,A)$, $j+if$ is in $S$. I claim that the map $$\begin{align*} \operatorname{Hom}(C,A) & \longrightarrow S, \\ f & \longmapsto j+if \end{align*}$$ is a bijection.

Indeed, the injectivity follows from the fact that $i$ is left-cancellable, and the surjectivity is due to the exactness:

If $k$ is in $S$, then $q(k-j) = \operatorname{id}_C - \operatorname{id}_C = 0$, which means $$ \operatorname{im}(k-j) \subseteq \ker q = \operatorname{im} i, \tag1 $$ and then there exists $f$ in $\operatorname{Hom}(C,A)$ such that $k-j = if$. Explicitly, given $x$ in $C$, by $(1)$ we know that there exists $y$ in $A$ such that $(k-j)(x) = i(y)$, and this $y$ is the unique element of $A$ with that property (since $i$ is injective); so we define $f(x) = y$.

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Write $$0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0 . \tag{1}$$

For each splitting $s : C \to B$ define $$\phi_s : A \oplus C \to B, \phi_s(a,c) = f(a) + s(c) .$$ It is well-known that $\phi_s$ is an isomorphism. Let $p_C : A \oplus C \to C$ denote the projection and $i_A : A \to A \oplus C, i_A(a) = (a,0)$. We have $$g \circ \phi_s = p_C \tag{2}$$ since $g(\phi_s(a,c)) = g(f(a) + s(c)) = g(f(a)) + g(s(c) = 0 + c = c$. Similarly $$\phi_s \circ i_A = f \tag{3}$$ since $\phi_s(i_A(a)) = \phi_s(a,0) = f(a) + s(0) = f(a)$. Now fix a splitting $\bar s : C \to B$. We get an isomorphism $\phi = \phi_{\bar s} : A \oplus C \to B$ satisfying $(1)$ and $(2)$. Clearly $\phi$ induces a bijection between splittings $s$ of $(1)$ and splittings $t$ of $$0 \to A \stackrel{i_A}{\to} A \oplus C \stackrel{p_C}{\to} C \to 0 \tag{4}$$ given by $s \mapsto \phi \circ s$. The splittings $t$ of $(4)$ have the form $$t(c) = (t'(c),c)$$ with a unique homomorphism $t' : C \to A$. Clearly the association

$$\operatorname{Hom}(C,A) \to \operatorname{Hom}(C,A \oplus C), t' \mapsto (t',id_C)$$ is a bijection.

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