I have been learning about modules in algebra for a few days now using Nicholson's Abstract Algebra, (I am a slow reader, it has taken me hours over the course of several days just to go through 7 pages) and I'm stuck on this small section discussing projections and projective modules:
If $_RW$ is free with basis $\{ w_1, \dots, w_n \}$, another important special case of Theorem 5 arises as follows: For each $k \in \{ 1, 2, \dots, n \}$ define $$ \pi_k: W \to R \quad\text{ by }\quad \pi_k(\Sigma r_i w_i) = r_k. $$ These are onto, $R$-linear maps (homomorphisms) for each $k$, called the projections associated with the basis $\{ w_1, \dots, w_n \}$. The following useful property is easily verified $$ x = \Sigma_k \pi_k(x) x_k, \text{ for all } x \in W. $$ In contrast to Corollary 1, a free module $W$ can only be the image of a module $M$ if $M$ contains a direct summand isomorphic to $W$. A module $P$ is called projective if it satisfies the following property: If $\alpha: M \to P$ is $R$-linear and onto then $M = \ker \alpha \oplus P_1$ for some submodule $P_1$ of $M$. Note that $P_1 \cong P$; indeed $P_1 \cong M / \ker \alpha \cong\alpha(M) \cong P$.
The Theorem 5 and Corollary 1 in question are:
Theorem 5: Let $_RW$ be free with a basis $\{w_1, \dots, w_n\}$. Given any module $_RM$ and arbitrary elements $x_1, \dots, x_n$ in $M$, define $$ \theta: W \to M \quad \text{ by } \quad \theta(\Sigma r_i w_i) = \Sigma r_i x_i $$ where the $r_i$ are in $R$. Then $\theta$ is an $R$-linear map such that $\theta(w_i) = x_i$ for each $i$.
Corollary 1: Every $n$-generated module is an image of a free module with a basis of $n$ elements.
I'm confused about a couple of things, especially in the last paragraph.
In contrast to Corollary 1, a free module $W$ can only be the image of a module $M$ if $M$ contains a direct summand isomorphic to $W$.
A free module $W$ can only be the image of $M$ under what map? I think it's $\theta$, because $\pi_k$'s codomain is a ring. But then this would presuppose that $M$ is free. Furthermore, it's not clear how it follows that $M$ contains a direct summand isomorphic to $W$. I don't even know why it's relevant.
The next confusion I have is simpler. Why does $P_1 \cong M / \ker \alpha$ follow from $M = \ker \alpha \oplus P_1$? I thought I saw a result that would justify this earlier in the section, but nope. There is nothing in the section nor in my notes that mention such a result.
I would greatly appreciate it if someone could clear the fog here!
Elaborating on Mariano's comment, the claim about free modules refers to any map.
Let $\phi: M \rightarrow W$ be a surjective homomorphism, where $W$ is a free module with basis $\{w_i\}_{i \in I}$. We have the short exact sequence:
$0 \rightarrow \ker \phi \rightarrow M \xrightarrow\phi W \rightarrow 0$
We will show that the sequence splits.
(If you don't know what that means, tell me and I will fill in the details)
Since $\phi$ is onto, we can choose, for each $i\in I$, an element $m_i \in M$ such that $\phi(m_i) = w_i$. Now, let's define $\psi: W \rightarrow M$ such that $\psi(w_i) = m_i$ (since $\{w_i\}_{i \in I}$ is a basis, choosing the image of the $w_i$ uniquely determines a homomorphism).
Now, observe that we have that $\phi \circ \psi = id_W$. After all, for each basis element, it follows from our definirion of $\psi$ that $\phi \circ \psi (w_i) = \phi(m_i) = w_i$.
Whenever we can find $\psi$ with this property, we say that the sequence above splits, and it follows that $M \cong \ker \phi \oplus W$. That is, $W$ is isomorphic to a direct summand of $M$.
(This is called the splitting lemma. Your book most likely has a proof of that. If you need help understanding it, feel free to ask.)
Now, it turns out that this property of every surjection "splitting" is very important. It is true for free modules, as we have shown, but it turns out that it holds for a wider class of modules, which we call projective modules.
As for your second question, try considering the homorphism $\beta: \ker \alpha \oplus P_1 \rightarrow P_1$ given by $\beta(k,p) = p$. Verify that $\beta$ is onto and that $\ker \beta = \ker \alpha$ and then use the first isomorphism theorem.