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}) $