Property of free modules

Question is regarding following property of free modules:

Let $P$ be a free $R$ module. To every surjective homomorphism $f:B\rightarrow C$ of $R$ modules and to every homomorphism $g:P\rightarrow C$ there exists a homomorphism $h:P\rightarrow B$ such that $g=f\circ h$

$$\begin{array}{ccccccccc} P\\ \downarrow{h} & \searrow{g} & \\ B & \xrightarrow{f} & C \end{array}$$

Please feel free to edit the diagram to make it look better.

Let $S$ be a basis for $P$. Take $a\in S$, we have $g(a)\in C$. As $f$ is surjective, we have $b\in B$ such that $f(b)=g(a)$. Define $h(a)=b$. We them have $g(a)=f(b)=f(h(a))=(f\circ h)(a)$. Extend this linearly to whole of $P$.

It remains to prove that the map is well defined.

Suppose that $a_i\in S$ such that $\sum_{i=1}^na_i=0$. Linear independece of elements of $S$ implies that $a_i=0$. So, $h(\sum_{i=1}^n a_i)=\sum_{i=1}^n h(a_i)=0$.

Please let me know if this justification sufficient enough to say that $h$ is well defined?

• I don't think the issue is to show $f$ is well-defined. Of course $f$ is well defined. The issue of well-definedness comes up when you define a function on a element that's a coset, by using a representative element of the coset - in which case you have to show it's independent of the representative. In your case you need to show $f$ is a homomorphism, so show it respects addition and multiplication, and I don't see how you've done that. So you must show $f(a+b)=f(a)+f(b)$ and $f(ab)=f(a)f(b)$. – Gregory Grant Jul 25 '16 at 18:33
• What are the $a_i$? – Bernard Jul 25 '16 at 18:33
• @GregoryGrant : I have done some changes. Let me know if this is correct. I guess you mean $f(ra)=rf(a)$ – user311526 Jul 25 '16 at 18:38
• @Bernard : I have done some changes. Let me know if it is clear now. – user311526 Jul 25 '16 at 18:39
• What the $a_i$ are is clear now, but you prove only that $h(0)=0$, not that $h$ is well defined. Furthermore, $a_i=0$ can't happen t=if it's a member of a basis. – Bernard Jul 25 '16 at 18:47

If you have a basis $S$ for $P$, we know that $\operatorname{Hom}(P,C)\simeq C^S$, and similarly $\operatorname{Hom}(P,B)\simeq B^S$.
Now if for each $s\in S$, you choose an element $b_s\in B$ such that $\;f(b_s)=g(s)$ (we're using the axiom of choice here if $S$ is not finite), the family $\;(b_s)_{s\in S}\in B^S$ defines a homomorphism $h$ from $P$ to $B$, such that $f\circ h=g$ on $S$, hence on $P$.
• Well, what you should prove is taken for granted ($h$ well defined). It is equivalent to the initial assertion. Do you want me to add details about this fact? – Bernard Jul 25 '16 at 18:50