Proving : existance of the inverse/ bijective

**arbolis** · May 10th 2009, 06:24 PM

I don't know how to prove the following : Let $f:A \to B$ be a function. Prove that $f$ has an inverse if and only if $f$ is bijective.
I'm lacking of any idea about proving $\Rightarrow$ and $\Leftarrow$ .
I'm asking for a little push in the right direction and not the full answer. Thanks a lot.
P.S.: I got assigned to do this in a vector calculus class, if it helps.

**Jhevon** · May 10th 2009, 06:38 PM

Originally Posted by arbolis

I don't know how to prove the following : Let $f:A \to B$ be a function. Prove that $f$ has an inverse if and only if $f$ is bijective.
I'm lacking of any idea about proving $\Rightarrow$ and $\Leftarrow$ .
I'm asking for a little push in the right direction and not the full answer. Thanks a lot.
P.S.: I got assigned to do this in a vector calculus class, if it helps.

as stated, your problem seems to have nothing to do with vector calculus. the notation used is for real valued functions. anyway, it is pretty straight forward. just jump into it and use the definitions.

(=>) Assume f has an inverse, say g(x). We need to show that f is 1-1 and onto.

(a) f is 1-1: Assume f(x) = f(y). Apply the inverse function to both sides to get x = y. so f is 1-1.

(b) f is onto. Assume y is some element of B. Then it is an element of the domain of f inverse. and so, we have g(y) = x for some x in A. but this means f(g(y)) = f(x) which means...

(<=) Assume f is bijective. You must show that there exists a function from B to A which has the properties of an inverse function.

can you finish up?

**arbolis** · May 10th 2009, 07:06 PM

Thanks a lot Jhevon. I'll try it probably tomorrow because I'm very busy for now.
If I can't finish this up I'll ask for help here. (It is not probable, but sure.)

**arbolis** · May 11th 2009, 08:10 PM

$\Rightarrow$ ) (a) :
I assume $f$ has an inverse, $f^{-1}$ . (It means that if $f(x)=y \Rightarrow f^{-1}(y)=x \bold{ \{ \text{green} \}}$ ).

Let $f(x)=y$ , so we have that $f^{-1}(y)=x$ by $\text{green}$ .
If $f(x)=f(z)$ with $x$ and $z \in \mathbb{A}$ then $f(x)=f(z)=y \Rightarrow f^{-1}(y)=x$ and $f^{-1}(y)=z \Rightarrow x=z$ so that $f$ is injective.

(b):
Let $y \in \mathbb{B}$ and $f(x)=y$ for some $x \in \mathbb{A}$ .
By $\text{green}$ we have that $f^{-1}(y)=x$ for some $x \in \mathbb{A}$ , but this says that $f(f^{-1}(y))=f(x)$ . In other words that $\forall y \in \mathbb{B}$ , $\exists x$ such that $f(f^{-1}(y))=f(x) \Leftrightarrow \forall y \in \mathbb{B}$ , $\exists x$ such that $f(x)=y$ hence $f$ is surjective.
Hence f is bijective.
$\Leftarrow$ ) :
Let f be bijective with $dom f= \mathbb{A}$ and $Im f= \mathbb{B}$ .
As $f$ is injective, $f(a)\neq f(b)\Rightarrow a\neq b$ .
As $f$ is surjective, $\forall y \in \mathbb{B}$ , $\exists$ $a \in \mathbb{A}$ such that $f(a)=y$ . Each element of $\mathbb{B}$ has an antecedent in $\mathbb{A}$ .
Hence there exists a function $g$ such that $g(y)=a$ and $g(z) \neq a \forall z \in \mathbb{B}$ and $z \neq y$ .
If $f(a)=y \Rightarrow$ $g(y)=a$ , which means $f$ is invertible. (Proof completed.)

I hope I didn't make any error.

**Jhevon** · May 12th 2009, 06:48 PM

Originally Posted by arbolis

$\Rightarrow$ ) (a) :
I assume $f$ has an inverse, $f^{-1}$ . (It means that if $f(x)=y \Rightarrow f^{-1}(y)=x \bold{ \{ \text{green} \}}$ ).

Let $f(x)=y$ , so we have that $f^{-1}(y)=x$ by $\text{green}$ .
If $f(x)=f(z)$ with $x$ and $z \in \mathbb{A}$ then $f(x)=f(z)=y \Rightarrow f^{-1}(y)=x$ and $f^{-1}(y)=z \Rightarrow x=z$ so that $f$ is injective.

this is correct. but it is an awkward way to set it up.

i would rather use the rule that $f^{-1}(f(x)) = x$ for all $x \in \text{dom}(f)$ .

so, $f(x) = f(y) \implies f^{-1}(f(x)) = f^{-1}(f(y)) \Longleftrightarrow x = y$ .

(b):
Let $y \in \mathbb{B}$ and $f(x)=y$ for some $x \in \mathbb{A}$ .

sorry, you are begging the question here. unless, you are stating what you want to prove here. in which case, you should begin the next line by: "Assume $y \in B$ ".

By $\text{green}$ we have that $f^{-1}(y)=x$ for some $x \in \mathbb{A}$ , but this says that $f(f^{-1}(y))=f(x)$ .

using the rule i mentioned earlier, you could stop here.

In other words that $\forall y \in \mathbb{B}$ , $\exists x$ such that $f(f^{-1}(y))=f(x) \Leftrightarrow \forall y \in \mathbb{B}$ , $\exists x$ such that $f(x)=y$ hence $f$ is surjective.
Hence f is bijective.

superfluous, i think, as i mentioned above.

$\Leftarrow$ ) :
Let f be bijective with $dom f= \mathbb{A}$ and $Im f= \mathbb{B}$ .
As $f$ is injective, $f(a)\neq f(b)\Rightarrow a\neq b$ .

this is false. this is not what injective means. in fact, this is the converse of what it means to be injective.

As $f$ is surjective, $\forall y \in \mathbb{B}$ , $\exists$ $a \in \mathbb{A}$ such that $f(a)=y$ . Each element of $\mathbb{B}$ has an antecedent in $\mathbb{A}$ .
Hence there exists a function $g$ such that $g(y)=a$ and $g(z) \neq a \forall z \in \mathbb{B}$ and $z \neq y$ .
If $f(a)=y \Rightarrow$ $g(y)=a$ , which means $f$ is invertible. (Proof completed.)

I hope I didn't make any error.

it seems you relied on your definition of injective to do this. so it is wrong.

for this direction, i would use the fact that a function is a relation. that is, we can think of $f$ as the set $\{ (a,b) \mid a \in A \text{ and } b \in B \}$ . now, you just want to show that the inverse relation, $\{ (b,a) \mid a \in A \text{ and } b \in B \}$ is also a function. that is, you have to show that each $b \in B$ appears as a first coordinate in a pair once and only once. this will happen provided $f$ is bijective

**arbolis** · May 13th 2009, 01:04 PM

Thank you Jhevon for your detailed answer.
I have 2 questions : For proving $\Rightarrow )$ , (b) : if I start by "Assume that $y \in B$ " and I follow as I did, do I prove that f is bijective?

Second question, I don't see why what I wrote is false :

As

is injective,

.

You said :

this is false. this is not what injective means. in fact, this is the converse of what it means to be injective.

When I think about an injective function, I can imagine a function which is strictly increasing or decreasing over the interval it is injective. So for me, taking 2 different values $f(a)$ and $f(b)$ means that these values have a different antecedent. I don't see where I make an error by doing so.
The definition of injective states that if $f(a)=f(b)$ then $a=b$ .
I think that in logic if A implies B, then ¬A implies ¬B which is what I think I've done. So if the former is true, so is the latter... or I'm mixing things up?

This (calculus I ?) problem arises in my vector calculus course in the chapter of inverse functions. Maybe it's just a review, but I'm strongly willing to fill my weaknesses on basic properties.

**Jhevon** · May 13th 2009, 06:44 PM

Originally Posted by arbolis

Thank you Jhevon for your detailed answer.
I have 2 questions : For proving $\Rightarrow )$ , (b) : if I start by "Assume that $y \in B$ " and I follow as I did, do I prove that f is bijective?

yes, what you wrote after would work. it does prove bijectivity. i think it can be refined though, as i directed

Second question, I don't see why what I wrote is false : You said : When I think about an injective function, I can imagine a function which is strictly increasing or decreasing over the interval it is injective. So for me, taking 2 different values $f(a)$ and $f(b)$ means that these values have a different antecedent. I don't see where I make an error by doing so.
The definition of injective states that if $f(a)=f(b)$ then $a=b$ .
I think that in logic if A implies B, then ¬A implies ¬B which is what I think I've done. So if the former is true, so is the latter... or I'm mixing things up?

in red is your mistake. A => B is NOT the same as ~A => ~B. (do the truth table for both, you will see they are not the same). A => B is equivalent to ~B => ~A. thus, to use not equal, you must show that $a \ne b \implies f(a) \ne f(b)$

(note that a parabola fulfills your definition, parabolas are not injective)

This (calculus I ?) problem arises in my vector calculus course in the chapter of inverse functions. Maybe it's just a review, but I'm strongly willing to fill my weaknesses on basic properties.

yes, it is a good review. take your time to get these concepts straight.

**arbolis** · May 13th 2009, 06:56 PM

Hey Jhevon, many thanks!
I'll take note of the exercise. I wasn't aware of my logic error. I'm very glad to know about this mistake...
I've many exams coming very soon and after them I'll try to redo this problem in a better way.
Just a very last question (for these days at least) : in post #2, you said :

(a) f is 1-1: Assume f(x) = f(y). Apply the inverse function to both sides to get x = y. so f is 1-1.

By the inverse function, do you mean I take $f^{-1}(f(x))=f^{-1}(f(y)) \Rightarrow x=y$ ?
If so, I saw it, but I didn't realize it proved f is injective!

**Jhevon** · May 13th 2009, 07:28 PM

Originally Posted by arbolis

Hey Jhevon, many thanks!
I'll take note of the exercise. I wasn't aware of my logic error. I'm very glad to know about this mistake...
I've many exams coming very soon and after them I'll try to redo this problem in a better way.

well, good. doing some more problems like these will help expose any other errors you may think are true.

try proving some other basic results about functions. for example, if f(x) and g(x) are bijective, then f(g(x)) is bijective.

things of that nature. give two functions arbitrary properties, and prove whether or not their composition possess the same properties.

Just a very last question (for these days at least) : in post #2, you said : By the inverse function, do you mean I take $f^{-1}(f(x))=f^{-1}(f(y)) \Rightarrow x=y$ ?
If so, I saw it, but I didn't realize it proved f is injective!

yes, in fact, that is what i did in post #5. it proves it is injective since we showed that f(x) = f(y) => x = y

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