Universal Property of Localization Let $R$ be a commutative ring with $1\not =0$, and let $D\ni 1$ be a multiplicative subset of $R$. Consider the universal characterization of $D^{-1}R$:
There is a morphism $\pi\colon R\to D^{-1}R$ such that for all rings and morphisms $\psi\colon R\to S$ satisfying


*

*$\psi(1)=1$

*$\psi(D)\subset S^{\times}$


there is a unique morphism $\Psi\colon D^{-1}R\to S$ such that $\Psi\circ\pi=\psi$.

Suppose $D^{-1}R\not=0$. Prove directly from the universal characterization that $\pi(D)\subset (D^{-1}R)^{\times}$.
Note: See p. 707 in Dummit and Foote

I've been looking at this exercise for some minutes now, and I'm deeply confused. Please correct me if I am wrong (and I am sorry for that), I'm only writing as an "answer" here because this might be too long for a comment.
I don't think you can solve this question purely by using the universal property given above.
Why is this?
There is a morphism $id: R \rightarrow R$ such that for all rings and morphisms $\psi: R \rightarrow S$ satisfying


*

*$\psi(1)=1$

*$\psi(D) \subset S^{\times}$


there is a unique morphism $\Psi: R \rightarrow S$ such that $\Psi \circ id = \psi$.
Hence from this point of view, we are in the same setting. However, the statement here is certainly not true.
(I know it's not a healthy reference, but the universal property given in the wikipedia article on the topic includes the property $\pi(D) \subset (D^{-1}R)^{\times}$)
