What does it mean for a function to be welldefined? I encountered with this term in an excersice asking to check if a linear transformation is welldefined.

6It usually means that you have to check that the definition provides a unique output value for each input value. – Stephen MontgomerySmith Dec 14 '13 at 19:51

1It means the function doesn't depend on the way an element of the domain set is expressed. For your particular example one should check what's the map and the vector spaces involved. – DonAntonio Dec 14 '13 at 19:51

3A function $f:A \to B$ is well defined if for all $a \in A$, $f(a) \in B$ – Amihai Zivan Dec 14 '13 at 19:51

9@AmihaiZivan No, that just means it is defined. – Thomas Andrews Dec 14 '13 at 19:53

@ThomasAndrews Adding "If $a=b$ then $f(a)=f(b)$" will complete the definition? – Amihai Zivan Dec 14 '13 at 19:56

@Ami: yes, but don't you think you should verify that for yourself? – The Chaz 2.0 Dec 14 '13 at 20:04

@The I was honestly trying to learn. – Amihai Zivan Dec 14 '13 at 20:05

Fair enough, but it looked like you were trying to *teach* above... =) – The Chaz 2.0 Dec 14 '13 at 20:06

Relevant: [How do I prove that a function is well defined?](http://math.stackexchange.com/q/313169) – MJD Dec 15 '13 at 03:39

@ThomasAndrews Can we say with certainty that it is defined? What if $A = \{{a, b}\} B = \{{0, 1}\} $ and $f = \{{(a,0), (a,1), (b,0)}\} $? The claim that for all $a \in A, f(a) \in B$ still holds, right? – Ardent Oct 31 '21 at 12:55

@Ardent i would like to know the answer to this. Is this considered defined? How does defined differ from well defined? – Oofy2000 May 18 '22 at 15:31
1 Answers
All functions are welldefined; but when we define a function, we don't always know (without doing some work) that our definition really does give us a function. We say the function (or, more precisely, the specification of the function) is 'welldefined' if it does.
That is, $f : A \to B$ is welldefined if for each $a \in A$ there is a unique $b \in B$ with $f(a)=b$.
This often comes up when defining functions in terms of representatives of equivalence classes, or in terms of how an element of the domain is written. For example, the 'function' $f : \mathbb{Z} \to \mathbb{Z}$ defined by $$f(n) = \text{the first digit of the decimal expansion of}\ n\ \text{after the decimal point}$$ is not a welldefined function: we get $f(1)=0$ and $f(0.999\dots)=9$, even though $0.999\dots = 1$. We could turn it into a welldefined function by saying that the chosen decimal expansion must not have recurring $9$s.
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4@AndrePoole: It depends how you define it... it is welldefined as long as you declare something like $$\mathtt{abs}(x) = \text{the greatest integer}\ \le x$$ This only depends on the *value* of $x$, not its representation as a number, so it's welldefined. – Clive Newstead Dec 14 '13 at 20:02


Very good example as well as the presentation of what makes a function/map welldefined  this is the way it was taught to me when I also came across this same predicament. – Procore Sep 10 '16 at 20:02

1@AndrePoole Note, that the inverse of statement 1 is not true: while for every a there is just one b (or: you only get one result when throwing a into the formula), this b could be the result of several (or *all*) a. b=a² is well defined, but b=1 is the answer to both a=1 and a=1. – Trish Nov 13 '17 at 09:07