Characterization of projective and injective modules In Theorems 4.7 and 8.4 Hilton & Stammbach give two lists of 5 different characterizations of projective and injective modules, respectively. Even though I can follow the proofs they give, I'd like to get rid of the characterization involving free and cofree modules (the fifth one). Why? Because I don't want to go through the trouble of defining cofree modules (please don't insist on this).
The theorems say that for a module $P$ the following properties are equivalent:
1) $P$ is projective.
2) The functor $\operatorname{Hom}_\Lambda(P,\,\cdot\,)$ is right exact.
3) For every epimorphism $\epsilon:B\to P$, there exists a morphism $\sigma:P\to B$ such that $\epsilon\sigma=1_P$.
4) $P$ is a direct summand in every module of which it is a quotient.
and that for a module $I$ the following properties are equivalent:
1') $I$ is injective.
2') The functor $\operatorname{Hom}_\Lambda(\,\cdot\,, I)$ is right exact.
3') For every monomorphism $\mu:I\to B$, there exists a morphism $\delta:B\to I$ such that $\delta\mu=1_I$.
4') $I$ is a direct factor (or summand) in every module which contains $I$ as a submodule.
I know how to show that $1\Rightarrow 2\Rightarrow 3\Rightarrow 4$ and $1'\Rightarrow 2'\Rightarrow 3'\Rightarrow 4'$. How would I show that $4\Rightarrow 1$ and $4'\Rightarrow 1'$? Or prove the equivalence in any other way.

For $4 \Rightarrow 1$ assume $P$ satisfies $4$ and let $M \twoheadrightarrow N$ be surjective with a map $P \to N$.  The pullback $M \times_N P \to P$ is surjective (you can prove this categorically or from the definition of the object using that $M \twoheadrightarrow N$ is surjective. Alternatively, see An Introduction to Homological Algebra by Rotman, Exercise 5.10 on page 227) so by $4$ it is just a projection to a summand isomorphic to $P$, say $M\times_N P \cong P\oplus Q$. Then the inclusion of that summand, and the other half of the pullback square ($P \to P\oplus Q \cong M \times_N P \to M \to N$) gives that $P \to N$ factors through $M$.  Thus $P$ is projective (I assume you're using the definition stating that $P$ is projective if given an epimorphism $ψ:M→N$ and any morphism $f:P→N$, there exists a morphism (not necessarily unique) $g:P→M$ such that $f=ψ\circ g$).
I suspect a dual argument using the pushforward works for $4' \Rightarrow 1'$, but I haven't thought about it.
Edit: Note that $i_P\colon P \to M \times_N P$ splits $\pi_2$, so $\pi_2\circ i_P = \mathrm{id}_P$.  Then $f = f\circ\pi_2\circ i_P = \phi\circ\pi_1\circ i_P$ so $f$ factors through $\phi$.
