# Bijection between splittings and homomorphisms

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!

• Sorry, I have edited the question. The question is for modules. Aug 28, 2022 at 15:22
• 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 $(B\to C)\circ s=id_C$).
– Dave
Aug 28, 2022 at 15:44
• 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!).
– Dave
Aug 28, 2022 at 16:03

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 $$f(x) = y$$.
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.