Proof:

We begin by proving (ii) implies (iii). Note that $\phi$ is injective and so im$\;\phi$ is isomorphic to $K$. Let $x \in G$ be arbitrary and observe that

$$x - \phi \circ g(x)$$

lies in the kernel of $g$ and hence in the image of $f$. Thus,

$$x = (x - \phi \circ g(x)) + \phi \circ g(x) \in$$   im$$f +$$   im$$\; \phi.$$

We leave it to the reader to check that the sum $($im$f +$   im$\; \phi)$ is direct. It is easy to show that (iii) implies (i). We now show that (i) implies (ii). Let $k \in K$ and choose any $x \in G$ such that $k = g(x)$. Define $\phi(k) = x - (f\circ \theta)(x)$. To check that this is well defined, suppose that $k = g(x^{\prime}) = g(x^{\prime\prime})$ for a pair of elements $x^{\prime}, x^{\prime\prime} \in G$. There exists $y \in L$ such that $x^{\prime} - x^{\prime\prime} = f(y)$. Applying $f\circ\theta$ to this equation we get

$$(f\circ \theta)(x^{\prime}) - (f\circ \theta)(x^{\prime\prime}) = f\circ\theta\circ f(y) = f(y) = x^{\prime} - x^{\prime\prime},$$

from which we see that $(f\circ \theta)(x^{\prime}) - x^{\prime} = (f\circ \theta)(x^{\prime\prime}) - x^{\prime\prime}$. It is trivial to see that the map $\phi$ that we have defined is a group homomorphism and satisfies the requirement $g \circ \phi =$   id$_K$. $\square$