# What are the implications of a surjection between free modules?

My professor said the following in his lecture:

Suppose $$M$$ and $$P$$ are free modules and $$\alpha: M\rightarrow P$$ is a surjection. Then $$\alpha$$ splits and the kernel of $$\alpha$$ is a summand of $$M$$.

I know that $$M$$ and $$P$$ being free means that they have a basis and hence, any element can be written as a linear combo of the basis elements. I also know that if $$\alpha$$ being split means that there exists a $$\beta: P\rightarrow M$$ such that $$\alpha\beta=1_{P}$$. I am having trouble understanding why this surjection of free modules implies that the surjection splits. How do I get started?

• Really this is because $P$ is projective, but it's even easier when it's free. Since $P$ is free we have a basis, so I can define a module homomorphism $P\to M$ just by sending the basis of $P$ wherever I want. So I can send the basis elements of $P$ to preimages under $\alpha$ since $\alpha$ is surjective.
– Dave
Feb 7 '21 at 23:36

The key property of the free module $$F$$ with basis $$\{f_i\}_{i\in I}$$ is the following:

For every module $$N$$ and every choice of elements $$n_i\in N$$, $$i\in I$$, there exists a unique module homomorphism $$\phi\colon F\to N$$ such that $$\phi(f_i)=n_i$$.

Let $$\{p_i\}_{i\in I}$$ be a basis for $$P$$. For each $$i$$, there exists $$m_i\in M$$ such that $$\alpha(m_i)=p_i$$. Define $$\beta$$ to be the (unique) morphism induced by mapping $$\beta(p_i)=m_i$$. Then $$\alpha\beta(p_i) = \alpha(m_i)=p_i$$. Thus, $$\alpha\beta$$ is the identity on the basis, hence is the identity morphism.

Now let $$A=\mathrm{ker}(\alpha)$$, and $$B=\mathrm{Im}(\beta)$$. If $$x\in A\cap B$$, then $$x=\beta(p)$$ for some $$p\in P$$, hence $$0 = \alpha(x) = \alpha\beta(p) = p$$. Thus, $$x=\beta(p)=\beta(0)=0$$, so $$A\cap B=\{0\}$$. And if $$m\in M$$, then $$\beta(\alpha(m))\in B$$, and $$m-\beta(\alpha(m))\in A$$, with $$m=\beta(\alpha(m)) + (m-\beta(\alpha(m))$$, proving that $$A+B=M$$. Thus, $$M=A\oplus B$$.

Note the last part does not use the facts that $$M$$ and $$P$$ are free. In general, if $$N$$ and $$N’$$ are any modules, and $$\alpha\colon N\to N’$$ splits (has a right inverse), then $$\mathrm{ker}(\alpha)$$ is a direct summand of $$N$$, and $$N$$ is isomorphic to a direct sum $$N’\oplus N’’$$.

The first part shows that free modules are projective. A module $$P$$ is projective if and only if, for every pair of modules $$M$$ and $$M’$$, and surjective module morphism $$f\colon M\to M’$$, if there is a morphism $$\alpha\colon P\to M’$$, then there is a morphism $$\beta\colon P\to M$$ such that $$f\beta=\alpha$$. That is: if you have a diagram $$\begin{array}{ccccc} && P \\ &&\downarrow\alpha\\ M&\stackrel{f}{\longrightarrow}&M’&\rightarrow &0 \end{array}$$ with bottom row exact, then there exists a morphism $$\beta$$ such that $$\begin{array}{ccccc} && P \\ &{\scriptstyle\beta}\swarrow&\downarrow\alpha\\ M&\stackrel{f}{\longrightarrow}&M’&\rightarrow &0 \end{array}$$ that makes the diagram commutative.

Free modules are always projective, and for some rings, projective modules are necessarily free; but for other rings, you may have projective modules that are not free. A module is projective if and only if it is a direct summand of a free module.