1
$\begingroup$

A function of one variable is called injective if $x_1 = x_2$ whenever $f(x_1) = f(x_2)$. I'd like to know how to talk about bivariate or multivariate functions that satisfy a similar condition.

For example, in the case of a bivariate function, $f : X \to Y \to Z$, we might have that $y_1 = y_2$ whenever $f(x,y_1) = f(x,y_2)$. Would one say "$f$ is injective in its second argument"? Or maybe "$f$ is injective in $Y$"?

UPDATE: To be clear, I'm asking about how to talk about $f$ without introducing variables such as $x \in X$ into scope. Notice that, in the univariate case, we can say "$f$ is injective", not mentioning any $x \in X$.

UPDATE#2: I can see now that the original question encouraged a more open-ended interpretation than I intended. To be even more clear, I'm looking for an answer of the form, "$f$ is P", where 'P' is an adjectival phrase that does not mention $f$. (When I asked for a "name for the property", this is what I was getting at.) Thus, "$f$ is injective in its second argument" fits. But "$f$ is such that, for all $x \in X$, $f(x,\cdot)$ is injective" does not.

UPDATE #3: I've come to realise that I don't know how to even ask this question intelligibly. In my UPDATE#2, I demanded that 'P' not mention $f$, yet an answer I claimed fit the bill, "injective in its second argument" implicitly mentions $f$ (via the word "its").

$\endgroup$
2
  • 1
    $\begingroup$ I would use injective in its second argument, or injective in $y$. Similarly, I've seen $f$ is (Lipschitz-) continuous in its second argument. $\endgroup$
    – B. Mehta
    May 27, 2017 at 22:08
  • $\begingroup$ One could say $f_x$ is a family of injective functions parameterized by $x$. $\endgroup$ May 28, 2017 at 2:55

2 Answers 2

2
$\begingroup$

The function $f(x,\cdot)$ is injective.

$\endgroup$
3
  • $\begingroup$ This answer is useful but troublesome to me because it brings $x \in X$ into scope. I think your answer says, about some fixed $x \in X$, that the function $g : Y \to Z$ defined by $g(y) = f(x,y)$ is injective. But I want to talk about the fact that this is true for all $x \in X$. $\endgroup$
    – m0davis
    May 27, 2017 at 23:42
  • $\begingroup$ In this answer, $x$ isn't quantified, so as long as you quantify $x$ appropriately, it's ok. For your note, sometimes people will write $g_x(y)=f(x,y)$ to point out that $g_x$ is a function of $y$ which depends on a fixed $x$. So you could say $g_x$ is injective for all $x$. $\endgroup$ May 28, 2017 at 0:22
  • $\begingroup$ @MichaelBurr, point well taken. I misinterpreted this answer. Nevertheless, it's still troublesome to me that the answer doesn't give what I'm calling a "name of the property". See my UPDATE#2. $\endgroup$
    – m0davis
    May 28, 2017 at 1:11
0
$\begingroup$

Why not define $g(y,x)=f(x,y)$? Given that you are implicitly currying the function in its definition (rather than defining it as a mapping over a 2-tuple), then certainly $g(y)$ is injective.

$\endgroup$
1
  • $\begingroup$ I'm not sure how to interpret what you're saying given that you've used $g$ as both a univariate and as a bivariate function. In any case, please see my UPDATE#2 for further clarification. $\endgroup$
    – m0davis
    May 28, 2017 at 1:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .