Prove that Phi^-1:W->V is an isomorphism

**Runty** · September 22nd 2010, 07:22 AM

Suppose $V$ and $W$ are vector spaces over a field $F$ .

Now suppose $\varphi :V\rightarrow W$ is an isomorphism. The inverse of this is defined as $\varphi^{-1}:W\rightarrow V$ . Prove that $\varphi^{-1}:W\rightarrow V$ is an isomorphism, by verifying that it is linear and bijective.

So far, this is what I have available.

$\forall w\in W, \exists v\in V$ such that $\varphi (v)=w$ . This is for the surjectivity of $\varphi$ .
Additionally, the aforementioned $v$ is unique, which is for the injectivity of $\varphi$ .
As such, I can define $\varphi^{-1}(w)=v$ .
This implies that $\varphi(\varphi^{-1}(w))=\varphi(v)=w$ and $\varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v$

My guess is that I should do the same things for $\varphi^{-1}$ , but just in reverse of $\varphi$ .
What I mean is:
$\forall v\in V, \exists w\in W$ such that $\varphi^{-1}(w)=v$ (surjectivity of $\varphi^{-1}$ ).
Since the aforementioned $w$ will be unique, this implies that $\varphi^{-1}$ is injective.
Both of these together would imply that $\varphi^{-1}$ is bijective.

I haven't gotten to the linearity part yet, as I'd like to check to make sure I'm on the right track, or if I'm assuming too much.

**Swlabr** · September 22nd 2010, 07:28 AM

Originally Posted by Runty

Suppose $V$ and $W$ are vector spaces over a field $F$ .

Now suppose $\varphi :V\rightarrow W$ is an isomorphism. The inverse of this is defined as $\varphi^{-1}:W\rightarrow V$ . Prove that $\varphi^{-1}:W\rightarrow V$ is an isomorphism, by verifying that it is linear and bijective.

So far, this is what I have available.

$\forall w\in W, \exists v\in V$ such that $\varphi (v)=w$ . This is for the surjectivity of $\varphi$ .
Additionally, the aforementioned $v$ is unique, which is for the injectivity of $\varphi$ .
As such, I can define $\varphi^{-1}(w)=v$ .
This implies that $\varphi(\varphi^{-1}(w))=\varphi(v)=w$ and $\varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v$

My guess is that I should do the same things for $\varphi^{-1}$ , but just in reverse of $\varphi$ .
What I mean is:
$\forall v\in V, \exists w\in W$ such that $\varphi^{-1}(w)=v$ (surjectivity of $\varphi^{-1}$ ).
Since the aforementioned $w$ will be unique, this implies that $\varphi^{-1}$ is injective.
Both of these together would imply that $\varphi^{-1}$ is bijective.

I haven't gotten to the linearity part yet, as I'd like to check to make sure I'm on the right track, or if I'm assuming too much.

A function is a bijection if and only if it has an inverse.

This map $\varphi$ is just a function. We know it has an inverse (as a function), which is just $\varphi$ . We also know that its inverse has an inverse,

$(\varphi^{-1})^{-1} = \varphi$ .

Thus, $\varphi^{-1}$ is a bijection because it has an inverse.

Do you understand that?

That said, you working did seem fine. It was just the long way round!

If you have any problems showing linearity, ask. But do try it first...

**Runty** · September 22nd 2010, 07:30 AM

Originally Posted by Swlabr

A function is a bijection if and only if it has an inverse.

This map $\varphi$ is just a function. We know it has an inverse (as a function), which is just $\varphi$ . We also know that its inverse has an inverse,

$(\varphi^{-1})^{-1} = \varphi$ .

Thus, $\varphi^{-1}$ is a bijection because it has an inverse.

Do you understand that?

I'd forgotten that bit about bijections. Thanks for reminding me. Now just to get to the linearity part.

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