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**Runty** Suppose $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle F$.

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

So far, this is what I have available.

$\displaystyle \forall w\in W, \exists v\in V$ such that $\displaystyle \varphi (v)=w$. This is for the surjectivity of $\displaystyle \varphi$.

Additionally, the aforementioned $\displaystyle v$ is unique, which is for the injectivity of $\displaystyle \varphi$.

As such, I can define $\displaystyle \varphi^{-1}(w)=v$.

This implies that $\displaystyle \varphi(\varphi^{-1}(w))=\varphi(v)=w$ and $\displaystyle \varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v$

My guess is that I should do the same things for $\displaystyle \varphi^{-1}$, but just in reverse of $\displaystyle \varphi$.

What I mean is:

$\displaystyle \forall v\in V, \exists w\in W$ such that $\displaystyle \varphi^{-1}(w)=v$ (surjectivity of $\displaystyle \varphi^{-1}$).

Since the aforementioned $\displaystyle w$ will be unique, this implies that $\displaystyle \varphi^{-1}$ is injective.

Both of these together would imply that $\displaystyle \varphi^{-1}$ 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.