I've looked at several posts here and proofs elsewhere and it is not clicking in my brain.

Let $R$ a ring and $P$ be a direct summand of a free module. That is, there exists a (left) $R$-module $Q$ such that $F=P \oplus Q$ is free. Then, let $B= \{b_i\}$ be a basis for $F$. Further suppose we have an epimorphism $f:A \to B$ and a homomorphism $g:P \to B$.

Here is the bit that confuses me. Since $g$ is an epimorphism we can find $a_i \in A$ such that $f(a_i)=g(b_i)$ for each $i$. $g$ is only a map from $P \to B$ not from $F \to B$. How are we able to make a consideration like this? Why does it work out?

From there, define $\bar{h}:F \to A$

$$\bar{h}(\sum r_ib_i)=\sum r_im_i$$

We can check that this map is well-defined and what not and restricting to $P$ gives the desired result.


1 Answer 1



You can use that the Hom functor commutes with direct sums: we have a canonical isomorphism $$\DeclareMathOperator{\Hom}{Hom} \Hom_R(P,X)\oplus\Hom_R(Q,X)\simeq \Hom_R(F,X), $$and we can associate to the homomorphism $g:P\to B$ the homomorphism $h=g\oplus 0:P\oplus Q\to B$.

  • $\begingroup$ So, some of those generating elements will be sent to 0? $\endgroup$
    – RhythmInk
    Oct 25, 2018 at 18:45
  • 1
    $\begingroup$ If they belong to $Q$, yes. Their $P$-componet will also have an image, possibly non-zero. But I don't think you have to worry with a basis of $F$. $\endgroup$
    – Bernard
    Oct 25, 2018 at 18:47

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