# well defined homomorphism

Let be $A$ and $B$ two rings and let $f$ be a "rule" that associates elements of $A$ to elements of $B$, but not necessarily in a unique way, so that $f$ is a multifunction.

If I want to show that $f$ is a well defined homomorphism, is it enough to verify the following four statements?

• $f(0)=0$
• $f(a+b)=f(a)+f(b)$
• $f(ab)=f(a)f(b)$
• $f(1_A)=f(1_B)$

The last three statements ensure that the multifunction behaves well respect the ring properties and the second statement with the first ensures that $f$ is indeed a function.

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Is is the inverse of an usual homomorphism? – guaraqe Jun 1 '12 at 11:08
I don't really understand the question. Why not check that $f$ is a function first, then check that it's a homomorphism, instead of trying to do both at once? – Qiaochu Yuan Jun 1 '12 at 14:02
@QiaochuYuan It kind of looks like the OP believes there are "well defined multivalued homomorphisms" other than regular homomorphisms. – rschwieb Jun 1 '12 at 14:03
@rschwieb: well, there are (math.stackexchange.com/questions/148715/…), but I interpreted "well-defined homomorphism" to mean "well-defined function which is a homomorphism." – Qiaochu Yuan Jun 1 '12 at 14:09
It's not clear to me what these equalities even mean when $f$ is a multifunction. – Chris Eagle Aug 9 '12 at 14:00

Let us suppose you verify those three properties, and $f(a)=b$ and $f(a)=b'\neq b$.
Then $0=f(a-a)=f(a)-f(a)=b-b'\neq 0$, a contradiction.
I think the OP means $f(a + b) = f(a) + f(b)$ in the sense of sets ... so $f(a) - f(a) = \{b - b' \mid b \in f(a), b' \in f(a)\}$, so you can extend. And how do you now that $f(a-a) = f(a) - f(a)$? For that would give $\{0\} = f(a) - f(a)$ and therefore that $f(a)$ is a singleton, or? – martini Jun 1 '12 at 11:10
@martini I could have misunderstood, but I couldn't see how the outputs could be sets of elements in $B$. The only interpretation I could apply was this one. – rschwieb Jun 1 '12 at 11:14