In Group theory proofs what is meant by “well defined”

What is exactly meant or required for a mapping to be well defined. I was reading the first Homomorphism theorem (link) and the first thing the proof does is define a map and find it if its well defined. Intuitively it makes sense, but what are the requirements for a map to be well defined? For example in the link given, I understand they show one-one relationship as being well defined and later on they again prove its injective.

What have I understood wrongly?

Soham

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When teaching this year I stumbled upon this post by Gowers. It was helpful to me, at least. – Dylan Moreland Sep 3 '12 at 20:51
The quickest explanation is that a "well-defined map" is a function. That is, the image of any given element in the domain, however you write or express it, is a single element in the range. This looks like showing one-to-oneness, but it's only half of that. – Kirk Boyer Sep 3 '12 at 20:54
I think we use this expression to mean that even though the definition involves making an arbitrary (and sometimes hidden) choice, we can check afterwards that the result does not depend on that choice. In the link, they define $\theta(gK) = f(g)$. By writing this, we implicitly chose a representative $g$ of the class, and use it in our definition. But as it turns out, any choice of representative yields the same result, so this is "well defined". – Joel Cohen Sep 3 '12 at 20:56
@KirkBoyer Yes, I would like to listen some more. can you please elaborate. My question is, if indeed a "well defined map" is a function i.e a one to one relationship then in the link I added why do they go on proving its one-one again at the end?? – Soham Sep 3 '12 at 20:58
@Soham : Being "well defined" does not qualify the map but rather the definition itself, it means "this definition is actually rigorous". – Joel Cohen Sep 3 '12 at 21:38

One interesting observation is that "well-defined" is basically the converse of (so closely related to) "one-to-one". That is:

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

We say that $\varphi$ is one-to-one if $\varphi(g)=\varphi(h)$ implies that $g=h$.

Thus, if we're trying to prove $\varphi$ is a one-to-one homomorphism (or perhaps even an isomorphism), we can sometimes get that $g=h$ if and only if $\varphi(g)=\varphi(h)$, using double implications the whole way, so that we simultanously prove that $\varphi$ is both well-defined and one-to-one, rather than dealing with them in two separate steps. That then leaves only showing homomorphism (and onto, if we're trying to prove isomorphism). It isn't always so simple--occasionally, we'll need a slick trick to show one-to-one, which doesn't neatly lend itself to reversal and showing well-defined. Still, it's a nice thing to keep in mind as a possibility.

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I think that was really helpful – Soham Sep 3 '12 at 21:10
What is the object $\phi$ about which you could ask the question "is this well-defined"? If it is a function, you are begging the question. If it is a relation, then notation like $\phi(g)$ doesn't seem to make sense. So maybe it is something else -- could you clarify what kind of object your definition of well-definedness applies to? – m_t_ Jun 30 '14 at 10:14
@mt_33: $\phi$ is a relation, given by a rule that putatively makes it a function. For example, one might attempt to define a function from the rationals to the integers by letting $$\phi\left(\frac mn\right)=m,$$ where $m$ is an integer and $n$ is a positive integer. But this definition doesn't actually give a function, since (for example) $$\phi\left(\frac24\right)=2\neq 1=\phi\left(\frac12\right).$$ We can avoid this issue (thereby making it well-defined) by further specifying that $\frac mn$ be in lowest terms. – Cameron Buie Jun 30 '14 at 11:20

Suppose that I try to define a map $f$ from $\Bbb Q$, the set of rational numbers, to $\Bbb Z$, the set of integers by setting $f\left(\frac{a}b\right)=a$; what is $f(1)$?

$1=\frac11$, so $f(1)=f\left(\frac11\right)=1$.

But wait! $1=\frac22$, so $f(1)=f\left(\frac22\right)=2$.

And $1=\frac{100}{100}$, so $f(1)=100$.

Obviously this doesn’t work: by my ‘definition’ $f(1)$ could be any integer at all. In other words, my supposed definition doesn’t actually define anything: $f(1)$ depends on which representation of $1$ as a fraction of two integers I use, and nothing in the ‘definition’ requires me to pick one particular representation. This supposed function is not well-defined.

On the other hand, every rational number $q$ can be uniquely represented in the form $\frac{a}b$ where $\gcd(a,b)=1$ and $b>0$. Had I defined $f(q)$ to be the numberator $a$ of this specific representation, $f$ would have been a genuine function: it would have been well-defined.

Checking that a mathematical object is well-defined is really just checking that it is defined: that the purported definition actually does unambiguously specify the object.

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It means "does not depend on choices made". Actually the "master" case for this is the following:

Let $A, B$ be groups and $N$ a normal subgroup of $A$. If $f:A\to B$ is a homomorphism with $N$ contained in its kernel, then there is a unique homomorphism $h\colon A/N\to B$ such that $h(xN)= f(x)$ for all $x\in A$.

To make the word "well-defined" appear, one reformulates this as follows: We want to define $h:A/N\to B$. Let $X\in A/N$ be an arbitrary element. Then there exists an $x\in A$such that $X=x N$. Set $h(X):=f(x)$. This is well-defined, i.e. it does not depend on the choice of $x$. For if also $x'N=X$ then $x'=x n$ for some $n\in N$, hence $f(x') = f(x)f(n)=f(x)$ because $N$ is in the kernel.

Note that some authors mistakenly use the term "well-defined" when they should really just say "defined" (as if saying "we defined an object and we did it well"). Please avoid this (ab)use of the term.

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Hmm, I don't recall ever seeing that abuse of the term. Maybe I have been fortunate in my choice of literature. – Harald Hanche-Olsen Sep 3 '12 at 20:51
@Hagen Though I do understand, but let me pose a question. In the example you quoted above, when is the mapping $h$ not well defined? – Soham Sep 3 '12 at 20:56
It would not be well-defined if $f(x')$ could differ from $f(x)$, that is if $N$ were not contained in the kernel of $f$. Hence not every homomorphism $A\to B$ induces a homomorphism $A/N\to B$, only those with suitable kernel. Actually, one could define a map by really chosing a fixed representative per coset, but that map would not be a homomorphism ... – Hagen von Eitzen Sep 3 '12 at 21:02
okay I think I am getting some idea...I will have to think and play with this in my mind – Soham Sep 3 '12 at 21:08
What would an example of the abuse of this terminology be? – Keenan Kidwell Sep 3 '12 at 21:13

Say you have an equivalence relation $\equiv$ which defines equivalence classes $[a]=\{b | b\equiv a\}$.

A function $F$, which is defined on elements, will be well-defined, as a function on the equivalence classes if $F(a)=F(b)$ whenever $a\equiv b$.

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This answer says it all. I would give it at least +2 if I could. – Lee Mosher Sep 5 '12 at 18:34

I often explain this via an analogy.

Imagine you have lots of oranges and you define a function that sends each segment of an orange to an apple. As you know, we may describe a particular orange by choosing one of its segments.

The question is; does this function necessarily make a function on whole oranges that agrees with the function on individual segments? Remember that the definition of function demands that each orange would have to be sent to a unique apple, we cannot have one particular orange being sent to two different apples.

The answer to the above is...not necessarily. The only way this could work is if after choosing two segments of the same orange you get the same output from the function.

This is exactly what is happening here, you may describe a particular coset by choosing one of its representatives. You have a function on the representatives and have defined a new "function" on cosets. Just like the above analogy, in order for this to be a true function on cosets we must check that after choosing two representatives for the same coset you get the same outputs.

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