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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.

Martin Thoma
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Dubious
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  • Is is the inverse of an usual homomorphism? – guaraqe Jun 01 '12 at 11:08
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    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 01 '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 01 '12 at 14:03
  • @rschwieb: well, there are (http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations/148730#148730), but I interpreted "well-defined homomorphism" to mean "well-defined function which is a homomorphism." – Qiaochu Yuan Jun 01 '12 at 14:09
  • @QiaochuYuan I think I'm interpreting it the same way as you, and that post is not what I meant. You can see what I mean below, that a map satisfying the properties he mentioned can't really be multivalued. Maybe there is something I overlooked? – rschwieb Jun 01 '12 at 14:16
  • @QiaochuYuan, you said "Why not check that f is a function first, then check that it's a homomorphism?" I agree with this but the point was to check this conditions with the four statements that I wrote. However the answer is negative! I'm sorry if my question was unclear, and I'm sorry also for the stupid question. – Dubious Jun 01 '12 at 14:59
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    It's not clear to me what these equalities even mean when $f$ is a multifunction. – Chris Eagle Aug 09 '12 at 14:00

1 Answers1

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The three things you are checking are inconsistent with strict multifunctions.

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.

By definition, a well-defined function is a multifunction with unique outputs (so not really a strict multifunction.)

rschwieb
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  • 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 01 '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 01 '12 at 11:14
  • @martini I'm working from this wiki. – rschwieb Jun 01 '12 at 11:23