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Sorry if this is a stupid question. This page on Math Is Fun uses the terms "general function", "surjective function", "injective function" and "bijective function". This seems to imply that a map can only be injective, surjective or bijective if it was already a function. Is this the case?

Take the example, $\mathbb{Q}\rightarrow \mathbb{Q}^{^{-1}}$. Since $0^{^{-1}}$ is undefined, not every element in the domain maps to an element in the codomain. Evidently, it's not a function.

But, since no number in the codomain maps to more than one number in the domain, can it still be an "injective map"?

Math is Fun says:

To be Injective, a Horizontal Line should never intersect the curve at 2 or more points.

A quick Desmos graph proves that to be the case for the map. Desmos graph showing the map

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    $\begingroup$ What is your definition of "map"? In my vocabulary there is no difference between "map" and "function". Sometimes I use "map" for functions that have special properties (e.g. continuity). $\endgroup$ – drhab Dec 13 '18 at 8:36
  • $\begingroup$ I think that's subjective. $\endgroup$ – Asaf Karagila Dec 13 '18 at 9:14
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Whether a mapping is a function can often be a matter of simply how you define two things:

  • How you define its domain
  • How you define the values the mapping takes on in said domain

For example, consider $f : \mathbb{R} \to \mathbb{R}$ as definned by $f(x) = 1/x$.

This is not a function as-is: $x=0$ presents an issue. However, we can fix this in two ways, either being valid:

  • We redefine the mapping in a sort of piecewise manner to take on a value at $x=0$; for example,

$$f(x) = \left\{\begin{matrix} 1/x & x \neq 0\\ 0 & x=0 \end{matrix}\right.$$

  • We redefine the domain and just exclude $0$ from it, i.e. instead say that the mapping is $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$

Boom, $f$ is defined for all values of the domain (and only maps each element of it to a single element obviously) and thus is a "general function" in the sense you noted.

Without either of these premises, at best $f$ is what we call a "partial function." Partial functions need not assign a value to each element of the domain. What this "general function" you were introduced to is really more often called is a "total function," where every element of the domain is assigned a value. Note that all total functions are partial functions: just because they don't have to assign a value to all of the domain, doesn't mean they can't.

Is there a notion of "injective" for partial functions? Yup!

Let us remember the definition of injective: for all $a,b$, $f(a)=f(b)$ implies $a=b$. In that sense, you could consider a partial function injective under the same definition provided $f$ is defined at those values.

So can the mapping be injective, even if not defined at all elements of the domain? Well ... yes. But it's just a slightly different notion of injective than it is for total functions. You could say it's injective but only in the sense in which partial functions are injective.

Similarly, you could define similar notions for "partial" surjectivity and in turn "partial" bijectivity - just include a premise that, wherever we have the mapping take on a value, it must also be defined there.

Beyond this, though, I have nothing to really offer you.

Why?

It's because it depends on the context whether you would say it's injective in the sense of "total function" injectivity, or in the sense of "partial function" injectivity. To say that total functions are just a subset of partial functions also carries with it that all "total injective" functions are "partial injective," but not all "partial injective" functions are "total injective" - because for a function to be "total injective" would mean it would have to be a total function in the first place, and not all partial functions are total functions.

You can probably see already how confusing a matter this is for me to approach, and it's probably not much better on your end, is it? I mean, the definitions for it are so similar that it muddies the waters a hell of a lot.

So context is important.

If, for example, this is being discussed in a homework assignment, and you've not been introduced to partial functions yet, then to say a function is injective in the "partial" sense and not the "total" sense is basically just a different way to say "it's not injective in the way that matters in this context," right? (Namely if the "function" isn't really a total function, it's definitely not a "total injective" function.)

Though of course you could also leave a note in said homework to the professor explaining why it's "partial injective" but not "total injective." I like doing that sometimes in cases where minor ambiguities or contradictions pop up just so my work has complete clarity, even if it's a minor detail.

So, in summary:

  • The mapping $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 1/x$ is not a "total function," but it is a "partial function." It could be tweaked to be a total function though.

  • The notion of "injective" for partial functions is the same as for total functions, with the slight change that when we assume $f$ is given a value, e.g. $f(a)$, it is assumed said value is defined. (We ignore undefined values.)

  • There are similar notions of surjective and bijective for partial functions.

  • Whether you want to say a partial function is injective depends on context. If you're under the expectation that the mapping in question be a total function to be injective, and you have a mapping that is injective but only a partial (not total) function, you won't say it's injective since you implicitly also want it to be total.

  • If you ever need clarifications on context and what is expected, ask your professor/teacher/whoever for the best clarity.

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  • $\begingroup$ Thanks very much! Perfectly informative. As to what you said about "context", I'm a year 9 (Australian version of third-year middle school I think) who has been given some extension work. Unfortunately, it's currently school holidays and thus, I can't really contact my teacher. He asked the question I used in my example and answered it by saying that since it wasn't a function it didn't qualify as surjective or injective. My answer was that it wasn't a function but it was bijective. I just wanted clarification because it was (as you said) all quite confusing. $\endgroup$ – PolymorphismPrince Dec 13 '18 at 9:15
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    $\begingroup$ Yeah, I didn't really encounter the notion of "total/partial" functions until this past couple of years of uni. So in your case, especially given the teacher's answer, it would be assumed you're working with total functions. (Thus "not function" basically means "definitely not an injective function.") But yeah it's confusing lol @_@ $\endgroup$ – Eevee Trainer Dec 13 '18 at 9:18

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