# 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 $P$.

• thanks for the answer but i do not see how this is different from what i have written.. – user311526 Jul 25 '16 at 18:46
• 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
• No No. please let me think about it. I thought i have proved it is well defined and other user said it is obvious. – user311526 Jul 25 '16 at 18:51
• I am confused now, what am i supposed to prove? When does well definedness case arises? when an element has two different representations. In this case no element has two different representations so there is no question of well definedness. Am i correct? – user311526 Jul 25 '16 at 18:53
• It is, iif you use the definition of a basis. The problem is that you speak of a basis, but don't really use its definition or its properties. – Bernard Jul 25 '16 at 18:56