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What is exactly meant or required for a mapping to be well defined? I was reading the First Isomorphism theorem (link), and the first thing the proof does is define a map and find out if it's 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 it's injective.

What have I understood wrongly?

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    $\begingroup$ When teaching this year I stumbled upon this post by Gowers. It was helpful to me, at least. $\endgroup$ Sep 3, 2012 at 20:51
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    $\begingroup$ 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. $\endgroup$
    – Kirk Boyer
    Sep 3, 2012 at 20:54
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    $\begingroup$ 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". $\endgroup$
    – Joel Cohen
    Sep 3, 2012 at 20:56
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    $\begingroup$ @Soham : Being "well defined" does not qualify the map but rather the definition itself, it means "this definition is actually rigorous". $\endgroup$
    – Joel Cohen
    Sep 3, 2012 at 21:38
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    $\begingroup$ Your post came to my attention as I was fixing broken links to PlanetMath.org. I think their topic page on proving the First Isomorphism Theorem is essentially the same as what you originally linked (the PlanetMath site went through a few reorganizations over the years). Have a look and let me know if you agree it is a suitable replacement. $\endgroup$
    – hardmath
    Feb 6, 2022 at 0:27

6 Answers 6

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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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    $\begingroup$ I think that was really helpful $\endgroup$
    – Soham
    Sep 3, 2012 at 21:10
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    $\begingroup$ 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? $\endgroup$ Jun 30, 2014 at 10:14
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    $\begingroup$ @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. $\endgroup$ Jun 30, 2014 at 11:20
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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 non-zero 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 numerator $a$ 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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    $\begingroup$ Hmm, I don't recall ever seeing that abuse of the term. Maybe I have been fortunate in my choice of literature. $\endgroup$ Sep 3, 2012 at 20:51
  • $\begingroup$ @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? $\endgroup$
    – Soham
    Sep 3, 2012 at 20:56
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    $\begingroup$ 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 ... $\endgroup$ Sep 3, 2012 at 21:02
  • $\begingroup$ okay I think I am getting some idea...I will have to think and play with this in my mind $\endgroup$
    – Soham
    Sep 3, 2012 at 21:08
  • $\begingroup$ What would an example of the abuse of this terminology be? $\endgroup$ Sep 3, 2012 at 21:13
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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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    $\begingroup$ This answer says it all. I would give it at least +2 if I could. $\endgroup$
    – Lee Mosher
    Sep 5, 2012 at 18:34
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    $\begingroup$ Yes for a beginner math student this is the easiest answer to understand and apply to problems. $\endgroup$ Oct 10, 2019 at 14:17
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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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Good-definiteness pops up in the agenda whenever you are trying to define a map from a quotient set to another set, say $\tilde f\colon (X/\sim)\to Y$. In this case the argument of the (candidate) map is an entire equivalence class, but there's no way to operatively dealing with equivalence classes without relying on a representative of their. So, while necessarily relying on some representative to define the map, you want this (arbitrary) choice to be irrelevant on the result of the map (tentatively) defined on classes as such. Therefore, for such a map to be "well defined" you need to prove that: $$[x']=[x]\Longrightarrow \tilde f([x'])=\tilde f([x]) \tag1$$ For example, for some map $f\colon X\to Y$, define the equivalence relation on $X$ given by: $$x'\sim x\stackrel{(def.)}{\iff}f(x')=f(x)$$ Then, the map $\tilde f$ defined by $\tilde f([x]):=f(x)$ fulfills the condition $(1)$ and hence is "well defined" (on $X/\sim$).

On the contrary, $\tilde f\colon (X/\sim)\to X$ defined by $\tilde f([x]):=x$ is not well defined, because for $x\ne x'\in [x]$ we have $[x']=[x]$, but $\tilde f([x'])=x'\ne x=\tilde f([x])$.

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