Free modules are projective.

Let $M$ be a free module. We have the diagram below (where the second row is exact)

enter image description here

Since $M$ is free, it has a basis (call it $X$). For every $x_i \in X$, there exists $b_i \in B$ such that $f(x_i)=b_i$. But since $g$ is surjective, there exists $a_i \in A$ such that $g(a_i)=b_i$. So I was planning to define the funtion $\bar{f} : M \rightarrow A$ by $\bar{f}(x_i) = a_i$ such that $f(x_i)=b_i$ and $g(a_i)=b_i$. But then I remembered that $g$ was not necessarily injective...so one element in $M$ can be sent to more than one element in $A$, right? And that would mean that $\bar{f}$ is not well-defined. So I was just wondering if there was a way to resolve this issue...

Thanks in advance.

  • 4
    $\begingroup$ How can one element of $M$ be sent to more than one element in $A$? You are taking the basis for $M$, and defining what $\bar{f}$ does to each $x_i$ and as you said choose any $a_i$ so that $g(a_i)=f(x_i)$, and by extending this linearly you get a map. If you are wondering if the different choices of $a_i$ yield different maps, and the answer is yes. The lift homomorphism needs not to be unique in the definition of projective. $\endgroup$ – Daniel Montealegre Oct 28 '13 at 21:55

There is no issue: for each $i$ you are choosing an $a_i\in A$ such that $g(a_i) = f(x_i)$. The definition of projective does not require the extension to be unique. You think of an example: consider $A = R^2, B = R$ and $g(r,s) = r + s$. Let $M = R$ and let $f:M\to B$ be the identity. Then there are many ways to extend $f$ - in fact infinitely many if $R$ is infinite.


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