Show that this functor is a localization 
Definition. Let C be a category, and $W \subseteq$ C a class of morphisms. A functor $F : \bf{C} \to D$ is said to be a
  localization of C with respect to W if
(i) $F(f)$ is an isomorphism for each $f \in W$, and
(ii) whenever $G : \bf{C} \to D'$ is a functor carrying elements of
  $W$ into isomorphisms, there exists a unique functor $G' : \bf{D}\to
D'$ such that $G'F = G$.

I want to show that the following functor $F$ is a localization. Let Ab the category of abelian groups and $W$ the class of morphisms
$f : A \to B$ such that $\ker(f)$ and $\mathrm{coker}(f)$ are torsion groups. Let D be the category
with the same objects, but with $Hom_{\bf{D}}(A,B) = Hom_{\bf{Ab}}(\Bbb{Q}\otimes A,\Bbb{Q}\otimes B)$. Define
$F : \bf{Ab} \to \bf{D}$ to be the functor which sends an object $A$ to itself and a map $f$ to $id\otimes f$.
What I've done so far:


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*(i): If $\ker(f)$ is torsion, then $\ker(F(f))=\Bbb{Q}\otimes \ker(f)=0$, so $F(f)$ is injective. To show that $\mathrm{coker}(F(f))=0$, I used that tensoring by $\Bbb{Q}$ is right-exact, so given the short exact sequence


$0\to A\xrightarrow{f} B\to \mathrm{coker}(f)\to 0$
gives the exact sequence
$\Bbb{Q}\otimes A\xrightarrow{id\otimes f} \Bbb{Q}\otimes B\to \Bbb{Q}\otimes\mathrm{coker}(f)\to 0$
from where I deduced that $\mathrm{coker}(F(f))=\Bbb{Q}\otimes\mathrm{coker}(f)=0$ since $\mathrm{coker}(f)$ is torsion. Hence, $F(f)$ is surjective, so it is an isomorphism. My first question is

Question 1: is my deduction from the exact sequence correct? I mean, does that exact sequence implies that $\Bbb{Q}\otimes \mathrm{coker}(f)$ is the cokernel of $id\otimes f=F(f)$?



*

*(ii):
Suppose thtat there exists $G:\bf{C}\to\bf{D}'$ is a functor satisfying (i). I have to show the existence of a unique functor $G':\bf{D}\to\bf{D}'$ such that $G'F=G$. 


Since $F$ is the identity on objects, I define $G'(A)=G(A)$ for every $A\in\bf{Ab}$ (and every such $G'$ should be defined that way on objects). On morphisms, we must have $G'(id\otimes f)=G(f)$, so my second question is

Question 2: Does this definition of $G'$ extend to every morphism in $Hom_{\bf{Ab}}(\Bbb{Q}\otimes A,\Bbb{Q}\otimes B)$?

 A: Question 1 : Yes you are correct. Note also that more is true : indeed $\mathbb Q$ is flat, that is, $\mathbb Q\otimes -$ is exact, so that the whole sequence $0\to \mathbb Q\otimes \ker(f) \to \mathbb Q\otimes A\to \mathbb Q\otimes B\to \mathbb Q\otimes \mathrm{coker} (f) \to 0$ is exact, so this also proves the previous claim that $\ker(F(f)) = \mathbb Q\otimes \ker(f)$. 
Question 2 : unfortunately no, not all maps $\mathbb Q\otimes A\to \mathbb Q\otimes B$ are of the form $\mathbb Q\otimes f$ for some $f$. For instance, take $A= \mathbb Q$ and $B= \mathbb Z$. Then there is only the zero map $A\to B$ (exercise), but there are many maps $\mathbb{Q\otimes Q\to Q}$ (both are isomorphic to $\mathbb Q$), so you have to proceed a bit differently. 
If you want a hint, un-hide the following; if you don't, keep it hidden. 

 Actually define $\mathbf{D}$ to be the category of abelian groups $A$ such that the canonical map $A\to A\otimes \mathbb Q$ is an isomorphism, that is, torsion free divisible groups, and $F(A) = A\otimes \mathbb Q$. Show that it is well-defined. And then you can define $G'$ on $\mathbf{D}$ as a restriction of $G$. Then show that my $\mathbf{D}$ is equivalent to yours. 

