Is a vertical line injective?

Question

Pretty much just the title. Obviously a vertical line isn't a function (doesn't pass the vertical line test), but technically it passes the horizontal line test for injectivity. But, I thought that an injective map needed to have distinct outputs corresponding with distinct inputs, which a vertical line doesn't have (multiple y outputs for our x input). Or does an injective map also have to be a function by definition?

Answer 1

As you noticed, a vertical line is not a function and therefore it is meaningless to discuss injectivity for it, which is defined using precisely the concept of map, that is

$$f(a)=f(b) \implies a=b$$

In some sense we could also claim that vertical line is surjective but also this statement is meaningless for the same reason.

Refer also to

• What exactly is a function?

Answer 2

Injectivity like surjectivity is a property of functions and as you've noted the vertical line is not the graph of a function. That being said, I can define a function which has the image of a vertical line by a function from $\mathbb{R}$ to $\mathbb{R}^2$ parametrically by $x = c, y = t$ with $c$ a real constant and $t$ a real variable. In this context the line would be the image of an injective function. We can instead use $y=kt$ for some real $k\neq 0$ and this parameterization will also be an injective function with the same image, the vertical line through $x=c$.

Answer 3

I think it depends on what you define "injective" to mean on something that is not a function.

It could be argued that "injective" talks about mappings and if something isn't a function it makes no sense to talk about it being injective. Fair enough.

But you could also say that if $F \subset X\times Y$ then if $F$ such that if $(a,w)\in F$ and $(b,w)\in F$ then that is only possible of $a=b$ is a valid definition of injective. So if vertical line is $F = \{(c,y)|y\in \mathbb R\}$ and if $(a,w)\in F, (b,w)\in F$ then $a=b=c$ regardless of the $w$ so the line is injective. Fair enought.

But one can say the latin for injective means into so that means we chart everything into. (This would be compatible with bijective= injective and surjective; it'd be hard to say a vertical line is "bijective" when for and $x \ne c$ we can't have any $(x,\sim)$.) So we need that for if there is no $(m,w)\in F$ for some $m \in X$ then $m$ isn't charted at all. so the definition for injection should include that that aspect. So as a subset $\{c\}\times \mathbb R$ the line is injective but as a subset of $\mathbb R\times \mathbb R$ it isn't. Note: for this definition of injective $F= \{(x,y)| x = \tan(y)\}$ will be injective but certainly not a function. (For any $x$ there will be a $y = \arctan x$ so $(x, y)\in F$. And if $(a, y),(b,y) \in F$ then $a = b =\tan y$. But it's certainly not a function as $(x, \arctan x)$ and $(x, \pi + \arctan x)$ are both in $F$ so we don't have the single input/single output condition.)

I'd say for all practical purposes "injection" only refers to a function and if something isn't a function forget about it. After all, why make trouble.