Splitting homomorphism implies splitting sequence Suppose $0\longrightarrow A\stackrel{\psi}{\longrightarrow} B\stackrel{\varphi}{\longrightarrow} C\longrightarrow 0$ is a short exact sequence of $R$-modules. Then $B=\psi(A)\oplus C'$ for some submodule $C'$ of $B$ with $\varphi(C')\cong C$ if and only if there is a homomorphism $\lambda :B\rightarrow A$ such that $\lambda\circ\psi$ is the identity map on $A$. 
I am confused about one direction of the proof of this proposition, as given in Dummit and Foote. Namely, the authors state that if $\lambda$ is given, then we simply define $C'=\ker\lambda\subseteq B$. Why does this work? In particular, why can we write every $b\in B$ uniquely as $\psi(a)+b'$ for some $a\in A$ and $b'\in\ker\lambda$, and further, why is $\ker\lambda\cong C$?
 A: "Why can we write...?":
If $b \in B$, then $b = b - \psi(\lambda(b)) + \psi(\lambda(b))$. Now we get $$\lambda(b - \psi(\lambda(b))) = \lambda(b) - \lambda(\psi(\lambda(b))) = \lambda(b) - \lambda(b) = 0,$$ such that $b -\psi(\lambda(b)) \in \text{ker}(\lambda)$. Therefore $b = b - \psi(\lambda(b)) + \psi(\lambda(b)) \in \text{ker}(\lambda) + \text{im}(\psi)$. 
Why is that unique?: 
Let $b = \psi(a) + \tilde{b} \in \text{ker}(\lambda) + \text{im}(\psi)$ and $b = \psi(a') + \tilde{b}' \in \text{ker}(\lambda) + \text{im}(\psi)$. Then we get $$\psi(a) - \psi(a') +\tilde{b} - \tilde{b'} = \psi(a) + \tilde{b} - (\psi(a') + \tilde{b}') = 0.$$ Applying $\lambda$ yields $a - a' = 0$ as $\tilde{b}, \tilde{b'} \in \text{ker}(\lambda)$, such that $a = a'$. Since $a = a'$, we get $\psi(a) = \psi(a')$ and therefore $$\tilde{b} - \tilde{b'} = \psi(a) - \psi(a') +\tilde{b} - \tilde{b'} = 0.$$ Thus we also have $\tilde{b} = \tilde{b'}$. This means we just showed uniqueness.
"Why is $\text{ker}(\lambda) \cong C$?":
Denote the restriction $\varphi \mid_{\text{ker}(\lambda)} \colon \text{ker}(\lambda) \rightarrow C$ by $\phi$. We will show that $\phi$ is an isomorphism. As $\varphi$ is surjective, any element $c \in C$ has a preimage $b = \psi(a) + \tilde{b}$ (we showed that every $b \in B$ can uniquely be written like that). That means we have $$\varphi(\tilde{b}) = \varphi(\psi(a)) + \varphi(\tilde{b}) = \varphi(\psi(a) + \tilde{b}) = c$$ by exactness. Thus $\phi$ is surjective as we found a preimage in $\text{ker}(\lambda)$. Now consider $b \in \text{ker}(\lambda)$, such that $\phi(b) = 0$, i.e. $b \in \text{ker}(\phi) \subset \text{im}(\psi)$. Thus it suffices to show that the intersection $\text{ker}(\lambda) \cap \text{im}(\psi)$ is trivial. Let $b \in \text{ker}(\lambda) \cap \text{im}(\psi)$. Then there exists $a \in A$, such that $\psi(a) = b$ and we also have $\lambda(b) = 0$. Thus $a = \lambda(\psi(a)) = \lambda(b) = 0$. This yields $b = 0$, such that $\phi$ is injective. This means $\phi$ is an isomorphism. 
