TITLE: Why are homomorphisms usually assumed to be functions?
QUESTION [7 upvotes]: A group homomorphism is usually defined to be a function $\phi$ such that if $x * y = z$, then $\phi x \times \phi y = \phi z$. However, this can be generalized. We could define that a group homomorphism is a relation $\phi$ such that if $x * y = z$ and $(x,x') \in \phi$, and $(y,y') \in \phi$ and $(z,z') \in \phi$, then $x' \times y' = z'$. This reduces to the usual definition in the case where $\phi$ is a function. However, I've never seen this generalization mentioned anywhere. Is there something wrong with it?
My thoughts are that it might have to do with functions having well-behaved preimages. In particular, if $f$ is a function then $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$. If $f$ were simply an arbitrary relation, this property could (would?) fail.
So my question is, why are homomorphisms usually assumed to be functions? Note that this is not a question about group homomorphisms specifically.

REPLY [2 votes]: Your proposed definition of homomorphisms as relations doesn't work when the relation $\phi$ isn't a function, at least if you make the reasonable assumption that $(e,e')\in\phi$, where $e$ and $e'$ are the identity elements of the two groups.  Suppose, for example, that some $x$ is related by $\phi$ to two different values, say $(x,x')\in\phi$ and also $(x,x'')\in\phi$.  Then, applying your definition with $y,y',z$,and $z'$ equal to $e,e',x$, and $x''$ respectively, we could infer from $x*e=x$ that $x'\times e'=x''$, which is wrong if $x'\neq x''$.
Instead of assuming that $(e,e')\in\phi$, you could deduce similar trouble from your definition plus the assumption that, for every $y$ in the domain group, there is at least one $y'$ in the codomain group with $(y,y')\in\phi$.
If you want to define homomorphic relations, you might reasonably define them to be subgroups of the direct product of the domain and codomain groups.