# Thoughts on this Inverse Bijection Proof

• blindgibson27
In summary: X, so g(f(x)) = z \in ZTherefore, g \circ f is onto Z.In summary, the conversation is discussing the concepts of one-to-one and onto functions and how to prove that a composition of two onto functions is also onto. The first part involves proving that the inverse of a one-to-one function is also one-to-one, and the second part involves proving that the composition of two onto functions is also onto.
blindgibson27

Is this sufficient?

#### Attachments

• Bijection Proof.png
4.3 KB · Views: 1,609
What are you trying to prove, exactly? These just look like definitions to me, in which case a much simpler description of a one-to-one function would be as follows: A function $f:X \rightarrow Y$ is an injection if, $\forall a,b \in X, \ f(a) = f(b) \implies a = b$.

It is to proof that the inverse is a one-to-one correspondence. I think I get what you are saying though about it looking as a definition rather than a proof.

Let $f:X\rightarrow Y$ be a one to one correspondence, show $f^{-1}:Y\rightarrow X$ is a one to one correspondence.

$\exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2}$

furthermore, $f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})$ (by definition of function $f$ and one to one)

kind of stumped from this point on..
I may want to transfer this post over to the homework section though, I did post to just get a confirmation on my thoughts on bijection but it is now turning into something a bit more specific than that

Your proof that $f^{-1}$ is injective is correct. Your proof that it is surjective does not look to me like it actually says anything!

You want to prove that, if $x\in X$ then there exist $y\in Y$ such that $f^{-1}(y)= x$.

Given $x\in X$, let $y= f(x)$. Then it follows that $f^{-1}(y)= f^{-1}(f(x))= x$.

That makes a lot of sense and I am following that thought process, thank you for clearing that up for me. I was just stumped on the direction of the onto.

In the same light, these are my thoughts on my next exercise. If I have this wrong I may need to solidify my idea on the concept a bit more.

It reads: Show that if $f:X \rightarrow Y$ is onto $Y$, and $g: Y \rightarrow Z$ is onto $Z$, then $g \circ f:X \rightarrow Z$ is onto $Z$Prf
Given $y \in Y$, let $y = g^{-1}(z)$ and $x = f^{-1}(y)$
$\forall z \in Z$, $f^{-1}(g^{-1}(z)) = f^{-1}(y) = x$

## 1. What is an inverse bijection?

An inverse bijection is a mathematical function that has a one-to-one correspondence between two sets, where each element in one set has a unique corresponding element in the other set. This means that the function has both an input and output, and for every input, there is only one possible output.

## 2. How is an inverse bijection different from a regular bijection?

An inverse bijection is the reverse of a regular bijection. In a regular bijection, every input has a unique output, but in an inverse bijection, every output has a unique input. This means that the roles of the input and output are reversed in an inverse bijection.

## 3. What is the significance of proving an inverse bijection?

Proving an inverse bijection is important because it confirms the existence of a one-to-one correspondence between two sets. This can be useful in various mathematical proofs and applications, as well as in understanding the relationship between two sets of data.

## 4. What are the steps involved in proving an inverse bijection?

The steps involved in proving an inverse bijection include showing that the function is both one-to-one and onto. This means that every element in the domain has a unique corresponding element in the range, and that every element in the range has at least one corresponding element in the domain. Additionally, the proof may involve using the properties of the function, such as composition and inverse functions, to show that the correspondence is valid.

## 5. Can an inverse bijection exist between two infinite sets?

Yes, an inverse bijection can exist between two infinite sets. As long as there is a one-to-one correspondence between the elements of the two sets, the inverse bijection can be proven. This is true even if one set is larger than the other, as long as the elements in both sets can be paired up in a one-to-one manner.

Replies
5
Views
990
Replies
6
Views
2K
Replies
8
Views
2K
Replies
7
Views
1K
Replies
1
Views
1K
Replies
2
Views
10K
Replies
14
Views
3K
Replies
9
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K