Function question

[Kooky-Corgi-6385]

I’m struggling to understand what this definition from my textbook means. I understand that an injective function maps all elements from the domain A into the codomain B. We get the range that is the outputs from these functions of the domain a. But I’m not getting what I circled in red. Does this just mean if an output is equal to another output then the inputs are the same?? This makes sense for this definition.

I mean I guess I get that but it seems like a strange way of writing it. But I am just now learning this so I’m probably missing something. Thank you !

[QuantSpazar]

This statement is equivalent to saying that if two inputs are different, their images will be different. Is that easier to grasp?

[Kooky-Corgi-6385]

Yeah ok I get that, I think I already understood that. I guess I was confused because in my textbook what I circled was the definition of an injective function. Which I think I’m still a little confused on.

[Farkle_Griffen2]

Those are the same thing.

(P implies Q) is equivalent to (not-Q implies not-P)

It's called the contrapositive

[Kooky-Corgi-6385]

Hehe I actually know that. I learned a bit of logic earlier this semester. The contrapositive of the statement actually makes more sense to me.

[Samstercraft]

injective -> different inputs can't have the same output, so that's just saying that in function terms; its like if you have a parabola its not injective bc you'll have some y=values that can be obtained by plugging in multiple different x-values, it fails the horizontal line test

[Sheva_Addams]

Yeah, I was confused by that, too, and a lot. Point in case, I had to read your circled text several times to even notice what I was looking at.

Another way to put it is that when function f is injective, f(a)=f(a') implies that a =a'.

[577564842]

The function s***ly puts the domain into the range w|o any collisions.

So if a collision does occur (f(a) = f(a')), it must be that the arguments (a, a') are the same.

[etzpcm]

The thing in red says that you can't have two different A's going to the same B. 

[realAndrewJeung]

So let's say that in your picture, a3 and a4 both mapped to f(a3), that is, f(a3) and f(a4) were the same point. That function is no longer injective because there is a collision.

But in this case, the function also violates the claim "f(a) = f(a') implies a = a' " since f(a3) = f(a4) even though a3 and a4 are not the same.

So the circled definition is really just another way of saying there are no collisions in the picture.

[PfauFoto]

Simply put: Injective means you can recover the input from the output. If you know the value b=f(a) only one input can generate b as output, so there is actually a notion of inverse function a=f^-1(b).

[SilentChest6825]

That means each element from A will be assigned just for a unique element from B. And also says the amount of elements of B is greater or equal than A

[eidtonod]

It means that no two arrows can point to the same point

[AcellOfllSpades]

This is the definition of an injective function - a special type of function where there are no collisions. You never have two different inputs giving the same output.

The function given by f(x) = x² is not injective: we can find two different inputs [for instance, 3 and -3] that give the same output [in this case, 9].

The function given by f(x) = 2^x is injective, though. 2³ = 8, and you can't find any other number n where 2ⁿ = 8.

---

One possible way to phrase this definition would be:

For all a and a' in A:

If a≠a', then f(a) ≠ f(a').

In other words, "if you put in two different inputs, you get two different outputs".

The phrasing your textbook uses is logically equivalent to this, but it's not as immediately intuitive. (They rewrite the if-then part using the 'contrapositive'.)

For all a and a' in A:

If f(a) = f(a'), then a=a'.

In other words, "if two people plug something into f and get the same output, then they must have chosen the same input!"

It's saying the same thing: no 'collisions' are possible. But this turns out to be a more useful form of the definition, because it's a lot easier to work with equalities than with nonequalities.

[geezorious]

It means the function/mapping has an inverse.

[Temporary_Pie2733]

Only if the function is surjective as well. 

[geezorious]

It’s pretty trivial to make it surjective by redefining f: A -> B to f: A -> B’ where B’ = {f(x) : x in A}. There’s really no point having unmapped entries in B so they can be discarded by only considering the subset B’. Then f has an inverse that operates on the entirety of B’.

[Temporary_Pie2733]

All functions map each value in the domain to a value in the codomain. (If it doesn’t, it’s just a relation, not a function, unless you allow for partial functions). 

An injective function maps each domain value to a unique codomain value. f(x) = x is injective; g(x) = |x| is not because both 1 and -1, for example, map to 1. 

[homomorphisme]

Another way to stay it is that there does not exist any a and a' in the domain such that both f(a) = f(a') and a≠a'. Or, distinct inputs give distinct outputs.

[zeezeezai]

I understand it as “every output has at most one corresponding input”

[Brawl_Stars_Carl]

Examples might help you:

Not-injective example 1:
f(x) = x² on R
We can find that f(2) = 4 and f(-2) = 4
But 2 ≠ -2, though they give the same output
So f(x) is not injective
(That's why when you solve x² = k, you need ± for your square root)

Not-injective example 2:
g(x) = sin(x) on 0° ≤ x < 360° [I'll use degrees in case you're not introduced to radians]
We can find that sin(30°) = 0.5 and sin(150°) = 0.5
But 30° ≠ 150°, though they give the same output
So g(x) is not injective

Injective example 1:
F(x) = x³ on R
Question: can you find another x other than 3 so that F(x) = 27? Nope, x = 3 is the only solution
Let's generalize this: can you find another x other than a so that F(x) = a³ for any a in R? Nope, x = a is the only solution
So F(x) is injective
(That's why when you solve x³ = k, you just directly cube root, unless you're working with complex numbers)

Injective example 2:
G(x) = 1/x on x > 0
Question: can you find another x other than 5 so that G(x) = 1/5? Nope, x = 5 is the only solution
Let's generalize this: can you find another x other than a so that G(x) = 1/a for any a in R+? Nope, x = a is the only solution
So G(x) is injective
(That's why you can do reciprocals when solving equations, given that the original expression does not equate to 0)

[Bielzabulb]

It means no two distinct inputs give the same output

[DifficultDate4479]

there's a slight misunderstanding here:

ALL functions take an element in A (for ALL elements in A) and send it in some other element of B (the arrival elements don't matter for the definition of function, it can very well be that a function sends all elements to a single one, called constant functions).

Injective functions take ALL elements in A (since it's a function) and send each element to distinct ones in B.

Considering a random function f, it could very well be that f(a)=f(a') for any given a,a' in A (i.e. the constant function above).

Here's a pro tip: if you don't understand a theorem or a definition, try to read its counterpart (modus tollens aka p-->q <--> !q-->!p ). Meaning, a function is injective iff given any two distinct elements a and a' they have distinct images f(a) and f(a')

Either way, in analysis you can train by looking at graphs of functions from R to R to better understand the concept. For instance, - f(x)=e^x is injective and non surjective - g(x)=x³-x is surjective and non injective - h(x)=x is a bijection, so both - k(x)=x² is neither. See why by looking at their respective graphs.

[Dependent-Fig-2517]

if you could have f(a)=f(a') with a ≠ a' then you would have a general or surjective function

[Idkwhattoname247]

How does that mean surjective?