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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.

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Number Nine

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blindgibson27

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How about this..

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

[itex] \exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2} [/itex]

furthermore, [itex]f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})[/itex] (by definition of function [itex]f[/itex] 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

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HallsofIvy

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Homework Helper

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

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

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blindgibson27

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blindgibson27

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It reads: Show that if [itex]f:X \rightarrow Y[/itex] is onto [itex]Y[/itex], and [itex]g: Y \rightarrow Z[/itex] is onto [itex]Z[/itex], then [itex]g \circ f:X \rightarrow Z[/itex] is onto [itex]Z[/itex]Prf

Given [itex]y \in Y[/itex], let [itex]y = g^{-1}(z)[/itex] and [itex] x = f^{-1}(y)[/itex]

[itex]\forall z \in Z [/itex], [itex]f^{-1}(g^{-1}(z)) = f^{-1}(y) = x[/itex]

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.

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.

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.

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.

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.

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