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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.

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    It's not clear to me what these equalities even mean when $f$ is a multifunction.2012-08-09

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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.)

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    @mart$i$ni I'm working from [this wiki](http://en.wikipedia.org/wiki/Multivalued_function).2012-06-01