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Before the immediate responses come in, I realize that a properly defined function means that it is defined for every value in its domain.

My question is this: if $f:A\to B$ has the property $f(a)=b_1$ and $f(a)=b_2$, then it is often still called a function, but one which is "not well-defined".

If there is $b$ in $B$ such that there is no pre-image under $f$ then we say $f$ is "not surjective".

So what would we call a "function" which has the property that $f(a)$ is not defined for some $a$ in $A$? It seems like there should be a word for this, other than just saying $f$ is not a function.


Edit: I realize that a function which is not well defined is not actually a function. I'm talking about informal speak, for example in class how we say "let's check if this function is well defined" as though it were a function even if it weren't well defined. I'm wondering if there is an analogous phrase for maps which aren't defined on their whole domain. This is all informal, which is why I tagged it a soft question.

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  • $\begingroup$ Could you give an example of something called a "function" $f: A \to B$ with $f(a) = b_1$ and $f(a) = b_2$? $\endgroup$ – Krijn Aug 10 '15 at 1:36
  • $\begingroup$ That's why I used quotes. I'm just referring to the fact that we use the phrase "not well-defined function" informally, even though such a function doesn't exist since it is not well defined. You'd think there would be a counterpart for functions which are not defined for their whole domain $\endgroup$ – Elliot G Aug 10 '15 at 1:38
  • $\begingroup$ In measure theory, sometimes we talk about almost everywhere-defined functions. $\endgroup$ – fonini Aug 10 '15 at 7:11
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You're looking for partial function. A partial function $f$ from $X$ to $Y$ is a function $X' \to Y$, where $X'$ is some subset of $X$.

However, regarding your comment about functions that are not well-defined: there is no such thing as a function that isn't well-defined. If $f$ from $X$ to $Y$ is not well-defined, then $f$ is not actually a function, but only a relation (some subset of $X \times Y$). Partial functions are not considered functions either. The term "function" requires that for every input there is exactly one output.

What probably confused you is that we often define a function, and right after defining it check that it is well-defined. What we are really checking though, is that what we claimed was a function was in fact a function; that's why we call it well-defined, meaning our definition was not faulty. It is somewhat of an abuse of words to define something as a function before checking that it is well-defined; one should really first define it as a relation, and then prove the proposition that it is a function.

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  • $\begingroup$ Thanks. And I guess I should clarify that I. Realize that a function which isn't well defined is not a function, but am wondering if there is a similar word fortress functions (which would also not be functions). That's why I tagged it a soft question, since this isn't really about definitions. $\endgroup$ – Elliot G Aug 10 '15 at 1:33
  • $\begingroup$ math.stackexchange.com/posts/1391494/revisions nailed it, glad you got the tick. Also @goblin yes, I didn't upvote yours because it's like "Yeah, the OP is gonna totally follow this answer" $\endgroup$ – ToolPurger Aug 10 '15 at 1:42
  • $\begingroup$ Okay, I think goblin was just disagreeing with my notation $X \to Y$ for things that aren't functions. I edited that out. I think the answer is free of errors now. $\endgroup$ – 6005 Aug 10 '15 at 1:43
  • $\begingroup$ cough, cough.."multi-valued functions in the complex domain"... $\endgroup$ – Thorsten S. Aug 10 '15 at 12:09
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Its not completely clear what you're looking for, but:

Definition 0. Let $Y$ and $X$ denote sets. Then a relation of type $Y \leftarrow X$ is, by definition, a subset of $Y \times X$.

Relations form a locally-posetal dagger category that is sometimes denoted $\mathbf{Rel}$.

Explicitly:

Given relations $q : Z \leftarrow Y$ and $p : Y \leftarrow X$, their composite is defined as follows:

$$(q \circ p)(z,x) \iff \mathop{\exists}_{y:Y}(q(z,y)\;\&\; p(y,x))$$

Given a relation $r : Y \leftarrow X$, its converse $r^\dagger : X \leftarrow Y$ is defined as follows:

$$r^\dagger(x,y) \iff r(y,x)$$

Definition 1. Suppose $r : Y \leftarrow X$ is a relation. Then:

  • $r$ is said to be entire iff for all $x \in X$, there is at least one $y \in Y$ with $r(y,x)$.

  • $r$ is said to be deterministic iff for all $x \in X$, there is at most one $y \in Y$ with $r(y,x)$.

These can be characterized in a "quasi-algebraic" manner:

  • $r : Y \leftarrow X$ is entire iff $r^\dagger \circ r \geq \mathrm{id}_X$

  • $r : Y \leftarrow X$ is deterministic iff $r \circ r^\dagger \leq \mathrm{id}_Y$

A deterministic relation is also called a partial function. Entire relations are sometimes called multifunctions. So a "function" that gives multiple outputs $b_1,b_2$ at some input $a$ can be referred to as a multifunction.

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  • $\begingroup$ Do you have a source for this? (I don't doubt you, I just want to know what book!) $\endgroup$ – ToolPurger Aug 10 '15 at 1:43
  • $\begingroup$ @ToolPurger, Peter Freyd has a book "Categories, Allegories" that covers this kind of stuff (bit of a tough read though.) $\endgroup$ – goblin Aug 10 '15 at 1:47
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It is not possible that $f(a) = b_1$ and $f(a) = b_2$ unless $b_1=b_2$. However, we can consider functions that output sets, so $f(a) = \{b_1,b_2\}$ and $f(c) = \{b_3\}$. If we did that we could think of $f$ as taking multiple values on $a$ but only a single value on $b$. This would be called a multivalued function (https://en.wikipedia.org/wiki/Multivalued_function). E.g. inverses of trigonometric functions.

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  • $\begingroup$ Is a multivalued function $f: X \to Y$ very different from a function $f: X \to 2^Y$ where $2^Y$ is the powerset of $Y$? $\endgroup$ – Krijn Aug 10 '15 at 1:38
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    $\begingroup$ @Krijn, functions $X \rightarrow 2^Y$ turn out to be the same as relations $X \rightarrow Y$. (Think about why.) $\endgroup$ – goblin Aug 10 '15 at 1:49
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$\text{Partial function}$ - when some items in the domain map to nothing (for example $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f:x\mapsto\frac{1}{x}$ is technically a partial function because it's undefined at $0$)

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  • $\begingroup$ already upvoted you! :) $\endgroup$ – 6005 Aug 10 '15 at 1:34
  • $\begingroup$ Both downvoters, why? $\endgroup$ – ToolPurger Aug 10 '15 at 14:59
  • $\begingroup$ +1 for definition of partial function (the two downs are spurious....) $\endgroup$ – Alec Teal Aug 10 '15 at 16:31

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