Determining a rule which assigns each ring a unit element Probably this is easiest, but as I am somehow stuck I would be pleased about some comments.
What I give myself is a rule $f$ which does the following:
To every commutative ring $A$ with $1$ the rule $f$ assigns a unit $f(A)\in A^*$, and this assignment shall satisfy the following property:
If $\varphi: A\rightarrow B$ is any ring homomorphism, then $\varphi(f(A)) =f(B)$.
I am convinced that the only rules which can do this are the following two:
(i) $f(A)=1$ for each ring $A$ simultaneously,
(ii) $f(A)=-1$ for each ring $A$ simultaneously.
How could one show this (if it is right)?
 A: Given the context of the problem, I am assuming that you want to impose the following conditions:


*

*"Commutative ring" means commutative ring with identity.

*"Ring homomorphism" means a homomorphism that preserves identity elements.


In this case, your guess is correct.  Here is a proof.
First, consider that $f(\mathbb{Z}) \in \mathbb{Z}^* = \{1,-1\}$.  
Next, let $A$ be any commutative ring.  Then there is a unique homomorphism $\phi \colon \mathbb{Z} \to A$, determined by $\phi(1) = 1$.  Given your requirements for the assignment $f$, it follows that $$f(A) = \phi(f(\mathbb{Z})).$$  If $f(\mathbb{Z}) = 1$, we conclude from the equation above that $f(A) = 1$ also. If $f(\mathbb{Z}) = -1$, it follows similarly that $f(A) = -1$.  Thus $f$ falls into either case (i) or case (ii), to use your notation.
(How did I produce this answer?  I looked for a universal example where I could use concrete reasoning.  The commutative ring $\mathbb{Z}$ is "universal" in the sense that it has a unique homomorphism to every ring.  The problem became much easier after understanding this specific example!)
A: One program that often works for these sorts of questions is:


*

*Select one or more rings for which it is easy to solve for all possibilities

*Study if and how these solutions extend to all rings.

