Verifying universal property of Grothendieck group I'm trying to verify the universal property of the Grothendieck group. Let $\overline{C}$ be the set of isomorphism classes of finitely generated $R$-modules over say a Noetherian ring $R$ and let $C$ denote the set of all f.g. $R$-modules. Let $A$ be the free abelian group over $\overline{C}$. Let $S$ be the set $S = \{ [M] - [M^\prime] + [M^{\prime \prime}] \mid 0 \to M \to M^\prime \to M^{\prime \prime} \to 0 \text{ exact}, [M], [M^\prime], [M^{\prime \prime}] \in \overline{C} \}$. Let $K = A / \langle S \rangle$. 

Added: 
The universal property $(K, f)$ has to satisfy is the following (according to my current understanding): For any abelian group $G$ and additive function $\lambda : C \to G$ there exists a unique group homomorphism $h: K \to G$ such that $h \circ f = \lambda$.
In a diagram:
$$\begin{matrix}
 C &\xrightarrow{f} & A / \langle S \rangle = K \\
 \left\updownarrow{=}\vphantom{\int}\right. & & 
\left\uparrow{i^\prime_m}\vphantom{\int}\right.\\
C &\xrightarrow{\lambda} & G \end{matrix}$$
Sorry, I don't know how to draw diagonal arrows in latex, if anyone knows how to fix this please go ahead.

Now I want to show that for any abelian group $G$ and any additive function $\lambda : C \to G$ there exists a unique group homomorphism $h$ such that $h \circ f = \lambda$ where $f : C \to A / \langle S \rangle$. 
So I define a function $h: K \to G$ as $[M] + \langle S \rangle \mapsto \lambda (M)$. What I'm stuck with is: how do I show that $h$ is a group homomorphism that is, $h(a + b) = h(a) + h(b)$? It does not follow from the definition since $\lambda$ is any function. But I need it to show that $h$ is well-defined. 
Added
What I have so far: 
If $[M] + \langle S \rangle, [N] + \langle S \rangle \in K / \langle S \rangle$ then an element in $A$ mapping to $[M] + \langle S \rangle$ looks like $[M] + [P] - [P^\prime] + [P^{\prime \prime}]$ where the $P$s form an exact sequence. Similarly for $N$. Then 
$$ h([M] + \langle S \rangle + [N] + \langle S \rangle) = h \circ f (M + P - P^\prime + P^{\prime \prime } + N + S - S^\prime + S^{\prime \prime})
= \lambda (M + P - P^\prime + P^{\prime \prime } + N + S - S^\prime + S^{\prime \prime}) = \lambda (M) + \lambda (N)$$
How do we achieve that?
Now we want to do 
$$ \dots = \lambda (M + P - P^\prime + P^{\prime \prime }) + \lambda(  N + S - S^\prime + S^{\prime \prime}) = \lambda (M) + \lambda(P) - \dots + \lambda(S^{\prime \prime}) $$
 A: We should define $h(\sum_ja_j[M_j]+\langle S\rangle):= \sum_ja_j\lambda([M_j]),$ for any element $\sum_ja_j[M_j]\in A,a_j\in\mathbb Z.$ To see that this is well-defined, suppose $\sum_ja_j[M_j]+\langle S\rangle=\sum_kb_k[N_k]+\langle S\rangle,$ so that $\sum_ja_j[M_j]- \sum_kb_k[N_k]\in\langle S\rangle.$ This implies
$$\sum_ja_j[M_j]-\sum_kb_k[N_j]=\sum_{i}c_i([M_i']-[M_i]+[M_i''])$$ 
is an identity in $A$ for some finite sum over generators in $S.$ Hence, 
$$h(\sum_ja_j[M_j]-\sum_kb_k[N_k]+\langle S\rangle)=h(\sum_{i}c_i([M_i']-[M_i]+[M_i''])+\langle S\rangle)$$ 
which implies by definition
$$\sum_ja_j\lambda([M_j])-\sum_kb_k\lambda([N_k])=\sum_{i}c_i(\lambda([M_i'])-\lambda([M_i])+\lambda([M_i''])).$$ 
Since the RHS is generated by short exact sequences and $\lambda$ is additive, we find that 
$$\sum_ja_j\lambda([M_j])-\sum_kb_k\lambda([N_k])=0,$$
or in other words
$$h(\sum_ja_j[M_j]+\langle S\rangle)=h(\sum_kb_k[N_k]+\langle S\rangle).$$
Thus $h:A/\langle S\rangle\to G$ is indeed well-defined. To see that $h$ is a homomorphism is now trivial from the definition of $h.$
