Firstly, a definition:

Definition 1: A function $\phi : X \rightarrow Y$ is one-to-one if $\phi(x_1) = \phi(x_2)$ only when $x_1 = x_2$.

Now the question:

Students often misunderstand the concept of a one-to-one function (mapping). You see, a mapping has a direction associated with it, from $A$ to $B$. It seems reasonable to expect a one-to-one mapping that carries one point $A$ into one point of $B$, in the direction indicated by the arrow. But of course, every mapping of $A$ into $B$ does this, and Definition 1 did not say that at all. With this unfortunate situation in mind, make as good a pedagogical case as you can for calling the functions in Definition 1 two-to-two functions instead.

Definition and question ever so slightly modified from A First Course in Abstract Algebra by Fraleigh.

So this is my thoughts: First of all, despite the fact that Definition 1 did not say this, I thought it could be more or less inferred. As in, there is a formal, and intuitive definition of a derivative, but they more or less mean the same thing. But, I guess if this were considered a two-to-two the best argument I could think of is given $x_1 = x_2 \in A$, and $\phi(x_1):=y_1 = \phi(x_2):=y_2 \in B$, then the "diagram", so to speak looks more or less like:

$$ x_1 \rightarrow y_1 \leftrightarrow y_2 \leftarrow x_2 $$

So there is a kind of a two-to-two symmetry, but to me, it is a very weak (couldn't we just cut out the middle anyway?) argument, and don't believe this is what the question is getting at. Can someone explain to me what most students don't understand?

  • $\begingroup$ @PeterFranek: How so? I am not really sure what you want cleaning up on? $\endgroup$ – Dair Jun 27 '14 at 7:18
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    $\begingroup$ I believe what they're after is this: "Two different points in $X$ are always sent to two different points in $Y$." (as opposed to "There are two different points in $X$ that are sent to the same point in $Y$."). $\endgroup$ – Arthur Jun 27 '14 at 7:26
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    $\begingroup$ Wow, this question and its solutions reaval more potential misunderstandings by students than I had been aware of! Thanks for asking it. +1 $\endgroup$ – rschwieb Jun 27 '14 at 13:49

Most students (I was one of them) thought incorrectly that one-to-one function means that for any x, f(x) can only have one value.

Injective seems to be a less confusing word.

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    $\begingroup$ The first time I saw injective, I thought it meant not surjective, so there can be other confusions $\endgroup$ – Henry Jun 27 '14 at 7:27
  • $\begingroup$ I'm still a little bit confused, so if I am understanding, "thought incorrectly ... f(x) can only have one value", $f(x_1)$ can give multiple values, but $f(x_2)$ must give different values than $f(x_1)$ (given $x_1 \neq x_2$)? But one-to-one, does not imply that given one $x$, $f(x)$ gives one output? $\endgroup$ – Dair Jun 27 '14 at 7:50
  • $\begingroup$ The definition of a function makes $f(x)$ give a unique value. However, there may exist $y$ different from x, such that $f(x)=f(y)$. For example, consider the absolute value function which has the same function value for 1 and -1. In the graph of a function, injective functions would be the "horizontal line" test. $\endgroup$ – JB King Jun 27 '14 at 7:59
  • $\begingroup$ OHH! I think I get it know, it's misleading because most people just think of a normal function for every f(x) there is a unique y, but it should actually be that plus doing a "horizontal line" test? So, while x^2 is a function, it is not one-to-one, because it fails the horizontal line test. Whereas x^3 is? $\endgroup$ – Dair Jun 27 '14 at 8:05
  • $\begingroup$ @JBKing: And thus, two-to-two, would mean that given a point, for any other arbitrary point it must not produce the same output! And in the case of x^2 it does, give x = 1, y = 1, let x_2 = -1, y = 1 thus it is not injective! $\endgroup$ – Dair Jun 27 '14 at 8:09

Searched and found this comment by Terrence Tao on a blog at this link:

Too long for a comment.

Terence Tao Says:

John H. Conway once observed that the common terminology “one-to-one” for injections is horribly misleading, as it suggests that the definition of an injection is “something that sends one input to one output” – which is instead the definition of a function, not an injective function. (The problem is that “one-to-one” is intended to be in contrast to “many-to-one”, but is often instead viewed by beginners as a contrast to “one-to-many”.) He proposed instead the more accurate terminology “two-to-two” (sending two distinct inputs to two distinct outputs). This hasn’t caught on, however (not surprising though, given the severe alliteration).

  • $\begingroup$ Whoa, so an injective function is not necessarily a function, in terms of function vs relation? That is misleading. My high school terminology deceives me. $\endgroup$ – Dair Jun 27 '14 at 7:30
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    $\begingroup$ @user667648 An injective function is definitely a function, but a function need not be an injective function... $\endgroup$ – Ted Jun 27 '14 at 7:32
  • $\begingroup$ @user667648 The issue is that people apply the definition of function to the term injective function because the phrase "one-to-one" can be misinterpreted. $\endgroup$ – Brad Jun 27 '14 at 7:34
  • $\begingroup$ If a function is defined on a single point, isn't it still injective? That seems to defeat the 'two-to-two' terminology. $\endgroup$ – Wonder Jun 27 '14 at 7:44
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    $\begingroup$ @Wonder I would call it vacuously true because there aren't two distinct points but either way I don't like the terminology. $\endgroup$ – Brad Jun 27 '14 at 7:46

Re: ``so an injective function is not necessarily a function, in terms of function vs relation? That is misleading.''

Your are absolutely correct: a relation may is injective. A relation R from A to B is injective when $R^{-1}\ o \ R$ is contained in the diagonal id$_A = \{(a,a): a\in A\}$ of $A\times A$ (the graph of the identity function on $A$).

The single-valuedness of a relation, being a partial function, is the dual property: $R$ is single-valued (a partial function, one defined on a subset of $A$) when $R \ o\ R^{-1}$ is contained in the diagonal id$_B=\{(b,b): b\in B\}$ of $B\times B.$ Duality is in that $R$ is replaced with $R^{-1}.$

Now replace $\subseteq$ with $\supseteq$ in the first definition to arrive at $R^{-1}\ o \ R\supseteq$ id$_A.$ A relation with this property is a correspondence. A function is a single-valued correspondence (full domain).

The dual notion of the preceding one, $R \ o\ R^{-1}\supseteq$ id$_B,$ defines a surjection.

  • $\begingroup$ Good job being point-less ... I mean point-free ;) $\endgroup$ – Musa Al-hassy Jun 28 '14 at 22:57

one-to-one, misunderstood: "for every x, there is only one y=f(x)". This is the definition of a function

one-to-one: "if you have two different Xs, their f value is different"

"two-to-two": if you pick two different Xs, and evaluate the function on each, you get two different Ys


I agree that one-to-one is confusing, and agree that two-to-two neatly captures the idea. But the existing vocabulary is so entrenched that even recommendations from luminaries like Terence Tao and John Conway won't change it.

I try to convince my students that mathematicians only define things after they understand what they are trying to define. You don't really learn a concept from its definition, you learn it from examples and counterexamples (easy to provide for various kinds of mappings). Once you understand the concept you can make sense of a formal definition, even if it's ugly and the concept isn't well named.

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    $\begingroup$ +1 for your second paragraph. Though for the record, there are those of us for whom "one-to-one" seems like a natural description of an injective function, because a non-injective function would be "two-to-one" or "many-to-one." I guess in my mind, "one-to-one" implies "(only one)-to-one." $\endgroup$ – David Z Jun 27 '14 at 17:20

As the other answers indicate it seems you correctly identified the source of confusion for most students and your proposed solution (two-to-two mapping) is really elegant I think. Nevertheless, since it is not in the answers above, here another way in which the term one-to-one can confuse students (read: confused me as a student).

Living in Holland I had been familiar with the concepts injective and surjective (under their Dutch names injectief and surjectief) for quite some time before starting to read math texts in English. Deducing meanings of words from context and similarities between languages is often easy enough and so the first so many times I encoutered the term one-to-one mapping I read it as [mapping establishing a] one-to-one correspondence. (In other words: a bijection.)

Only after the tenth or so instance of someone explicitly emphasizing that a map was onto in addition to being one-to-one, I figured that it might be a good idea to maybe look up the definitions of these terms...


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