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Why is it that math so focuses on the subclass of relations known as functions? I.e. why is it so useful for us in nearly all branches of mathematics to focus on relations which are left-total and left-unique? Left- (or even right-) totality seem to be intuitive, since if an element doesn't appear in the domain, we might throw it out. But why left-uniqueness?

I'm looking for something like a "moral explanation" of why they would be the most useful subclass of relations.

My apologies if this is a previous question; I looked and didn't find much.

MJD
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Eric Auld
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    I see $x \le y$ pretty frequently in math texts... Not to mention $x=y$. More seriously, I think that equivalence and order relations are pretty important, too, and the question is based on a false premise. You might try to sharpen it. – quid Dec 11 '14 at 01:01
  • @quid Point taken. – Eric Auld Dec 11 '14 at 01:07
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    Maybe related: http://mathoverflow.net/questions/121031/why-is-set-and-not-rel-so-ubiquitous-in-mathematics, http://math.stackexchange.com/questions/354779/the-category-set-seems-more-prominent-important-than-the-category-rel-why-is-th – Tyler Dec 11 '14 at 01:12
  • @quid: Hmm (binary) relations can also be considered as (boolean) functions on pairs. It really depends on which is considered more fundamental, doesn't it? – user21820 Dec 11 '14 at 03:33
  • @user21820 a function is a special type of binary relation, as explained in the question. The question asks why this special type of relation is so fundamental. And I replied that also other types of relations, such a order relations are quite relevant. Your remark seems neither here nor there. – quid Dec 11 '14 at 10:47
  • @quid: I mean that if you consider functions more fundamental, then you would see a relation as a special type of function and not the other way around. You don't since you already consider relations as more fundamental and hence consider functions as special cases. – user21820 Dec 11 '14 at 10:51
  • @user21820 OP starts with "Why is it that math so focuses on the subclass of relations known as functions?" It was not I who presented this point of view. – quid Dec 11 '14 at 10:57
  • @quid: Oh okay. I was just saying that this viewpoint is not necessarily the only one, and the other viewpoint may actually explain why we do special things for functions. For example, we write things like "$g(f(x))$" for functions $f,g$ and think of it as applying the two functions one after another to $x$, not in terms of any underlying relations. In fact, if one looks at the notation used, it is not so easy to define the function notation using local transformations in formal set theory. This suggests that our intuition of functions goes beyond it being just a special kind of relation. – user21820 Dec 11 '14 at 11:34

2 Answers2

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A function models a deterministic computation: if you put in $x$, you always get out the same result, $f(x)$, hence the left-uniqueness.

The asymmetry of the definition (left uniqueness rather than right uniqueness) is because the left side models the input and the right side models the output, and the input is logically prior to the output. If you know the input, you can determine the output, but you can't (in general) do the reverse. The function $f:x\mapsto x^2$ means that if you put in 17 you get out 289. But it makes no sense at all to ask what you get out before specifying what you put in.

MJD
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  • So you would say that the notion of time is an important factor even if the domain of the function has nothing to do with time? – Eric Auld Dec 11 '14 at 01:07
  • I want to say that your answer casts some light on why the notion of "locality" is asymmetric in domain and range. Like how a map between manifolds may be a bijective local diffeomorphism but have an image which not a submanifold, since it may double back. I feel like I'm describing something, but I'm not sure if I'm using the right words. – Eric Auld Dec 11 '14 at 01:11
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    There is a formulation of continuity in which the continuous functions are precisely the computable functions. Now consider a constant function as a prototypical continuous function. Viewed as a temporal process, it represents a complete destruction of all the information in the input, and information, once destroyed, cannot be recovered. So a constant function is continuous, but its inverse, in general, is not. – MJD Dec 11 '14 at 01:17
  • I like very much your implicit point that the loss or gain of information presupposes directionality. – Eric Auld Dec 11 '14 at 01:21
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I would say functions model our intuitive notion of directionality (this is a similar, though still slightly different, answer than MJD's). After all the notation for functions is $$X\xrightarrow{\;\;f\;\;}Y$$ and not $$X\overset{f}{\;\;\mathrel{\star\small\vdots\!\smallsmile\!\wr}\;\;}Y$$

Sal
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  • When talking about functions where the domain has nothing to do with time, why should we need a notion of directionality? Maybe a dumb question, but I think I could learn a lot from an answer. – Eric Auld Dec 11 '14 at 01:17
  • I'm saying the function has the property of being from $X$ to $Y$, and in that way has directionality, not that $X$ necessarily has any notion of "internal direction" or "internal time". – Sal Dec 11 '14 at 01:26
  • Right, why should we want such a notion of directionality? MJD above points out that it can refer to the gain or loss of information. – Eric Auld Dec 11 '14 at 01:28
  • Because it is a fundamental human notion. Why should we want a notion of the cardinality of a set? – Sal Dec 11 '14 at 01:38