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If $A\approx B$ then $A^{C}\approx B^{C}$ where $B^{C}:=\{f|f:C\to B\}$

by ''$\approx$'' I mean equinumerous

Proof: By hypothesis $A\approx B\Rightarrow \exists T:A\to B$ which is bijective and $\exists T^{-1}B\to A$ which is also bijective.

let now $\sigma: B^{C}\to A^{C}$ be defined as $\sigma: f\to T^{-1}\circ f$, for some $f\in B^{C}$

how to show that this function is bijective?

1-1

$\sigma(f_{1})=\sigma(f_{2})\Rightarrow T^{-1}\circ f_{1}=T^{-1}\circ f_{2}\Rightarrow f_{1}=f_{2}$ right?

surjective??

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2 Answers 2

up vote 2 down vote accepted

Your proof of injectivity is right, because you have $$T^{-1}\circ f_1=T^{-1}\circ f_2$$ Now, pre-compose both sides by $T$, to obtain $$T\circ T^{-1}\circ f_1=T\circ T^{-1}\circ f_2$$ But $T\circ T^{-1}$ is the identity function on $A$, hence you can conclude.

For surjectivity, let $f:C\longrightarrow A$ be an element of $A^C$. You need a pre-image of $f$ via $\sigma$, i.e. a function $g:C\longrightarrow B$ such that $\sigma(g)=f$. Take $g:=T\circ f$. To verify that this is the right one you only need to apply the definition of $\sigma$, so: $$\sigma(g)=\sigma(T\circ f)=T^{-1}\circ T\circ f=f$$ as claimed.

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Yes, your proof of injectivity is fine.

For surjectivity, take $f:C\to A$. Then let $f' = T \circ f$. What is $\sigma(f')$?

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