# Necessity of being well-defined in Group Homomorphism?

In Group Theory, homomorphism is isomorphism when we no longer restrict to bijective map; do we still need that map to be well defined in homomorphism (like in isomorphism) or homomorphism can be without requirement of "well-defined"?

P.S. We say that $\varphi$ is well-defined if $g=h$ implies that $\varphi(g)=\varphi(h)$.

• Yes, it still must be a well defined function (and it must respect the group structure, so $\varphi(g_1g_2) = \varphi(g_1) \varphi(g_2)$. – lulu Jul 27 '15 at 3:27
• You mean homomorphism, not homeomorphism. They are different things. The latter is from topology. You are putting way too much stock into the phrase "well-defined." There are no ill-defined maps, ever, period - only bona fide functions vs. failed attempts to define a function that didn't work out. – anon Jul 27 '15 at 3:28
• @anon - I haven't noticed for a long long time that it's homomorphism in group theory and homeomorphism in topology! Thanks – user231343 Jul 27 '15 at 3:31
• @Edi: I see; you might be interested in this question: math.stackexchange.com/questions/95741/… – coldnumber Jul 27 '15 at 3:38
• In reality, I would guess that many mathematics lecturers write things like "Define $f:G/N \to H$ by $f(gN) = h(g)$" (assuming that $h:G \to N$ is already defined), and then they say "First we must prove that $f$ is well-defined". Formally speaking this is wrong, because they haven't actually defined $f$ at all until they have proved it is well-defined, and it can sometimes be confusing to students. – Derek Holt Jul 27 '15 at 8:10

Functions, by their definition are well-defined: if one has a relation $f \in A \times B$, it is a function if and only if whenever $(a,b),(a,b') \in f$ we have $b = b'$. This allows to write (unambiguously) $b = f(a)$.

So functions can be:

1) One-to-one (these functions are called injective).

2) Many-to-one (for example, constant functions from a domain with more than one element).

Functions cannot be:

3) One-to-many. An image of a domain element must be a singleton.

However, with group homomorphsims, we often have homomorphisms defined on quotient groups (which are certain equivalence classes of a group).

Here is an example of something which is not well-defined: suppose we try to define-

$f: \Bbb Z_6 \to \Bbb Z$ by:

$f([a]_6) = a$. (Here, $[a]_6$ is the equivalence class of $a$ modulo $6$, where the equivalence is defined by:

$a\sim b \iff a-b$ is a multiple of $6$).

Note that $[3]_6 = [9]_6$, since $9-3 = 6$. But $f([3]_6) = 3 \neq 9 = f([9]_6)$.

We no longer have a unique image for $[a]_6$, it is not well-defined.

So when we define a function on a set of equivalence classes using elements contained in those equivalence classes, we need to be sure our choice of image is independent of the particular "representative" (element of the equivalence class) we calculate the image from. Another way of saying this, is that $f$ needs to be constant on each equivalence class, to induce a function on the equivalence classes.

Perhaps you have "well-defined" confused with injective, they look similar:

$f$ is injective if $f(a) = f(a') \implies a = a'$

Homomorphisms, unlike isomorphisms, are not required to be injective, isomorphisms must be (they must be surjective, or onto, as well). In fact, much of what is interesting about group theory comes from the fact that often, group homomorphisms are not injective, which leads to questions on "how much of the original group structure is preserved".