Currently I am reading through Dr. Weibel's The K Book.
There I encountered this exact sequence in the localization theorem of Heller:
$$K_{0}(\mathcal{B}) \longrightarrow K_{0}(\mathcal{A}) \longrightarrow K_{0}(\mathcal{A/B}) \longrightarrow 0,$$
where $\mathcal{B}$ is a Serre-subcategory of $\mathcal{A}$ and $\mathcal{A/B}$ is nothing but the localization of the abelian category $\mathcal{A}$ by the set $S$ where $$S = \{f \in\mathrm{morph}(\mathcal{A})\mid\ker f\text{ and }\operatorname{coker}f \in \mathcal{B}\}$$
Regarding this I have two questions:
1) Why is this exact sequence not left exact?
My reasoning : $\mathcal{B} \longrightarrow K_0(\mathcal{A})$ I can define an additive function $f$ just by taking the inclusion $\textbf{B} \in \mathcal{B}$ goes to [$\textbf{B}$]. Then by the universal property of the additive function I have a induced map from $$ K_0(\mathcal{B}) \longrightarrow K_0(\mathcal{A})$$ Which takes $[\textbf{B}]$ to $[\textbf{B}]$. So shouldn't it be?
2) In their proof they have taken $\Gamma$ to be the cokernel of $K_{0}(\mathcal{B}) \longrightarrow K_{0}(\mathcal{A})$
The aim is to show that $\gamma$ : $\mathcal{A/B} \rightarrow \Gamma$ is an additive function by defining $\gamma$$\textbf{(loc(A)) = [A]}$ where $\textbf{loc}$ is the exact functor from $$ K_{0}(\mathcal{A}) \longrightarrow K_{0}(\mathcal{A/B}).$$
To prove this we need to show for an arbitrary exact sequence in $\mathcal{A/B}$ $$0\rightarrow \textbf{loc\(A_0\)} \rightarrow \textbf{loc\(A_1\)} \rightarrow \textbf{loc\(A_2\)} \rightarrow 0$$ [$\textbf{A}_{1}$] = [$\textbf{A}_{0}$] $+$ [$\textbf{A}_{2}$] in $\Gamma$.
This is where I am stuck in the reasoning that they have given. I am attaching a photocopy of their proof of this part. I am confused especially in the part where they state that $\textbf{coker(g)} \in \mathcal{B}$ and $\textbf{loc(ker(g)} \cong \textbf{loc\(A_0\)}$
Sorry if the question is kinda long, but I am stuck here for a day, I am tahnkful for any help.
Q.1) Has a pretty good answer in the comments so I am posting the answer of my (Q.2) I think I figured it out, it suffices to show that in the following exact sequence if we can replace $loc(A_1)$ by $loc(A)$ and $j$ with $loc(g)$(defined later) we are done.
$$0\rightarrow loc(A_0) \rightarrow loc(A_1) \rightarrow^{j} loc(A_2) \rightarrow 0$$
Let it be the exact sequence in $\mathcal{A/B}$
Then $loc(A_1) \rightarrow^{j} loc(A_2)$ has a representation $$ A_1{\leftarrow ^{f}} A \rightarrow^g A_2$$ Where $f$ is a $\mathcal{B}$ iso.
But that implies that $loc(A) \cong loc(A_1)$
We have another exact sequence like this $$ 0\rightarrow ker(g) \rightarrow A \rightarrow A_2\rightarrow coker(g) \rightarrow 0 $$
Since $loc$ is an exact functor from $\mathcal{A} \rightarrow \mathcal {A/B}$ applying it on the above exact sequence we see that since $loc(A) \cong loc(A_1)$,
I have $loc(g)$ equivalent with the composition map $ j\circ \phi$ Where $\phi$ is the isomorphism from $loc(A)$ to $loc(A_1)$.
Proof : $\phi$ is represented by $$ A{\leftarrow ^{id}} A \rightarrow^f A_1$$
And we already know the the representation of $j$. So by the way they have defined the composition in the book we get $j \circ \phi$ as $$ A\leftarrow^{id} A\leftarrow^{id} A \rightarrow^{id} A \rightarrow ^g A_2$$
Which is nothing but $A\leftarrow^{id} A \rightarrow ^g A_2$ the map $loc(g)$