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Let $X,Y$ be sets and let $f\colon X\rightarrow Y$ be an invertible map.

We let $f^{−1}\colon Y\rightarrow X$ denote the inverse map of $f$.

For any permutation $\pi \in {\rm Perm}(X)$ of $X$ , define the map $\sigma\circ\pi\colon Y\rightarrow Y$ as the composite map $\sigma\circ\pi\colon=f\circ \pi \circ f^{-1}$. Thus, for any $y \in Y$ , one has $\sigma\circ\pi(y)= f\circ \pi\circ f^{-1}(y)$.

Show that $\sigma\circ\pi$ is a permutation of $Y$ .

Being invertible means it is bijective. so how do I go on from there?

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user271716
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1 Answers1

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It is enough to prove that $f\circ\pi\circ f^{-1}:Y\rightarrow Y$ has a left- and a rightinverse.

Functions that have a leftinverse are injective and functions that have a rightinverse are surjective.

The map $f\circ\pi^{-1}\circ f^{-1}$ "works" as left- and rightinverse of the map $f\circ\pi\circ f^{-1}$.

drhab
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  • sorry is it okay if i have a more detailed working on how to show this? because i have absolutely no clue how to go about doing it:( im sorry, really having difficulties with these. – user271716 Sep 26 '15 at 16:59
  • $\left(f\circ\pi^{-1}\circ f^{-1}\right)\circ\left(f\circ\pi\circ f^{-1}\right)=f\circ\pi^{-1}\circ\left(f^{-1}\circ f\right)\circ\pi\circ f^{-1}=f\circ\pi^{-1}\circ1_{X}\circ\pi\circ f^{-1}=f\circ\pi^{-1}\circ\pi\circ f^{-1}=\dots=1_{Y}$ and likewise $\left(f\circ\pi\circ f^{-1}\right)\circ\left(f\circ\pi^{-1}\circ f^{-1}\right)=1_{Y}$. This tells us that $f\circ\pi\circ f^{-1}$ is invertible, hence bijective. – drhab Sep 27 '15 at 06:53
  • The result of the previous problem shows that we have a well-defined map Σ : Perm(X)  > Perm(Y), π > σπ. Show that this map Σ is an invertible map. – user271716 Sep 27 '15 at 13:47
  • so for this question. do i assume that it is invertible? then i prove that both will get perm Y so means that it is invertible? since X -> X means that it is f o f^-1? – user271716 Sep 27 '15 at 13:48
  • I do not really understand your comment. Shortly it comes to: if $f:X\rightarrow Y$ is invertible then $\sigma_{\pi}:=f\circ\pi\circ f^{-1}:Y\rightarrow Y$ is also invertible. Its inverse is the map $\sigma_{\pi^{-1}}:f\circ\pi^{-1}\circ f^{-1}:Y\rightarrow Y$. That means exactly that $\sigma_{\pi}\in\text{Perm}\left(Y\right)$ since $\text{Perm}\left(Y\right)$ denotes the collection of invertible maps $Y\rightarrow Y$. You could also say that it denotes the collection of bijections $Y\rightarrow Y$, since a map is invertible if and only if it is bijective. – drhab Sep 27 '15 at 15:18
  • Show that this map Σ is an invertible map. so i just have to show that both will end up as the function perm Y? so to explain that Σ is invertible? – user271716 Sep 27 '15 at 15:24
  • no i have another question which comes from this question. The result of the previous problem shows that we have a well-defined map Σ : Perm(X)  > Perm(Y), π > σπ. Show that this map Σ is an invertible map. so how do i go about doing it? by the same method? – user271716 Sep 27 '15 at 15:31
  • What is $\Sigma$? "end up as the function permY..." $\text{Perm}(Y)$ is not a function. It is a collection of functions. Your wording is very sloppy. – drhab Sep 27 '15 at 15:31
  • If $\Sigma$ is a function on $\text{Perm}(X)$ prescribed by $\pi\mapsto\sigma_{\pi}$ then my answer proves that $\Sigma$ sends the elements of $\text{Perm}(X)$ to $\text{Perm}(Y)$. It can be proved $\Sigma$ itself is also invertible. Its inverse is the map prescribed by $\rho\mapsto f^{-1}\circ\rho\circ f$. – drhab Sep 27 '15 at 15:43
  • so i do the same method as just now to prove that it is invertible? sorry im having a lot of difficulties with this because its my first year in university and i just learnt this – user271716 Sep 27 '15 at 15:46
  • Yes. If $\tau_{\lambda}:=f^{-1}\circ\lambda\circ f$ for $\lambda\in\text{Perm}\left(Y\right)$ then $\tau_{\lambda}\in\text{Perm}\left(X\right)$ (same proof as for $\sigma_{\pi}\in\text{Perm}\left(X\right)$). Now let $T:\text{Perm}\left(Y\right)\rightarrow\text{Perm}\left(X\right)$ be prescribed by $\lambda\mapsto\tau_{\lambda}$. Then it can be proved that $\Sigma\circ T$ is the identity on $\text{Perm}\left(Y\right)$ and $T\circ\Sigma$ is the identity on $\text{Perm}\left(X\right)$. – drhab Sep 27 '15 at 15:58
  • your first part of the statement is for the inverse right? from perm Y to permX? – user271716 Sep 27 '15 at 16:10
  • I have stated that $\Sigma$ and $T$ are inverses of each other. Actually comments are not meant for this (we have gone too far allready), and I have decided to stop here. Just have a look at the stuff (especially my former comment), and if things remain unclear then pose another question on the subject. Eventually with reference to this question. Cheers. – drhab Sep 27 '15 at 16:18