The localization of a commutative ring with unity Let $S$ be a multiplicative set in a commutative ring $A$ with unity. We shall denote the localization of $A$ at $S$ with $A[S^{-1}]$.
Let $T$ be another multiplicative set in $A$ such that $S \subset T$. Further, let $j_S: A \to A_S$ be the canonical ring homomorphism defined by $j(a)=\frac{a}{1_S}$.
Claim: $A[T^{-1}]=A[S^{-1}][j_S(T)^{-1}]$.
To prove this claim, I have tried to write a rigorous argument by using the universal property of the localization of the ring. But I couldn't have done yet. Any help or suggestion?
 A: First, the universal property says that for any homomorphism $f: A\to B$ such that $f(S)\subset B^*$, the grup of units of $B$, there is a unique homomorphism $f': A[S^{-1}]\to B$ such that $f=f'\circ j_S$. 
Now, on one hand we have a homomorphism $jj: A[S^{-1}]\to A[S^{-1}][j_S(T)^{-1}]$ and the composition $h=jj\circ j_S$ send clearly the elements of $T$ to invertible elements. So by the universal property, there exist a unique homomorphism say $h':   A[T^{-1}] \to A[S^{-1}][j_S(T)^{-1}]$ such that $h=h'\circ j_T$. 
On the other hand, the homomorphism $j_T$ sends the elements of $S$ to invertible elements, so we have a map $j'_T: A[S^{-1}]\to A[T^{-1}]$ such that $j_T=j_T'\circ j_S$. This homomorphism sends the elements of $j_S(T)$ to invertible elements, so we have another homomorphism $ i: A[S^{-1}][j_S(T)^{-1}]  \to A[T^{-1}]$ such that $j_T'=i\circ jj$. 
Now, one only needs to show that $i\circ h'$ is the identity, as well as $h'\circ i$. But this is deduced from the unicity of the universal property.
For the first composition: from $h=jj\circ j_S$, $h=h'\circ j_T$,  $j_T'=i\circ jj$ and $j_T=j_T'\circ j_S$, we have $$j_T=(i\circ jj) \circ j_S=i\circ (jj \circ j_S)=i\circ h=(i\circ h')\circ j_T.$$
So $i\circ h':A[T^{-1}]\to A[T^{-1}]$ is a map that verifies the property of $(j_T)'$ corresponding to $j_T$ (it is different from $j'_T$, sory for the messing notation), which also the map $\operatorname{id}$ verifies. So by unicity it must be the identity. 
For the second composition $$h'\circ i: A[S^{-1}][j_S(T)^{-1}]  \to A[S^{-1}][j_S(T)^{-1}] $$
note first that $jj=h'\circ j'_T$, by its universal property, since 
$$h'\circ j'_T\circ j_S=h'\circ j_T=h=jj\circ j_S$$ and hence it verifies the same property that $jj$, which is uniquely determined by this property. But then 
we have $$jj=h'\circ j'_T=h'\circ (i\circ jj)=(h'\circ i) \circ jj$$
and we can argue as the first composition. 
