Let $\Phi:B\to A$ be a map of rings, $S\subset A$ a multiplicative subset, and $T\subset B$ a multiplicative subset of $B$ with $\Phi(T)\subset S$. Exercise 21.2.L of Vakil's Foundations of Algebraic Geometry claims that the pullback of differentials $\xi: S^{-1}A \otimes_A \Omega_{A/B}\to \Omega_{S^{-1}A/T^{-1}B}$ is an isomorphism. I am beginning to doubt this claim.
The previous exercise 21.2.K, which I believe I have completed correctly, asks us to precisely describe this pullback map in a general setting. I believe that chasing the construction gives us $\xi$ in the current situation as $a_1/s \otimes d_A a_2 \mapsto (a_1/s) d_{S^{-1}A} (a_2/1)$.
I do not see how to write down an inverse to this map. A generator of $\Omega_{S^{-1}A/T^{-1}B}$ must be of form $(a_1/s_1) d_{S^{-1}A} (a_2/s_2)$. I am bothered by where and how to send $1/s_2$. If, for instance, $\Phi(T)=S$ exactly, we know $(a_1/s_1) d_{S^{-1}A} (a_2/s_2)= (a_1/(s_1s_2))d_{S^{-1}A}(a_2/1)$ by Leibniz and triviality under pullback. So our construction for $\xi$ is automatically surjective and in this case we can easily write down the inverse to be the obvious thing. In general, the situation seems hopeless.
Is there something obvious that I am missing that allows one to see the claim in Exercise 21.2.L is in fact true?
It's best to break this problem up in to parts.
Step 1: prove that $S^{-1}\Omega_{A/B}=\Omega_{S^{-1}A/B}$. One good strategy is to define a derivation of $S^{-1}A$ in to $S^{-1}\Omega_{A/B}$ by sending $a/s\mapsto (1/s)da - (a/s^2)ds$, and then apply the universal property.
Step 2: prove that for any $T\subset B$ a multiplicatively closed subset which maps to invertible elements of $A$, we have $\Omega_{A/B}=\Omega_{A/T^{-1}B}$. This can be done by playing around with $d(a/s)$ - see here, for instance (do watch out for $A$ and $B$ swap places, though).
Don't hesitate to leave a comment if you're looking for an expanded argument.