a question about the dual space of a vector space: inverse of the dual is dual of the inverse? Suppose I have a linear/$\mathbb{R}$ map $T: V \rightarrow W$ invertible. 
I am wondering how can we show that $(T^{-1})^* = (T^*)^{-1}$? where $*$ denotes taking the dual. Any comments would be appreciated. 
 A: Let $V,W$ be vector spaces over a field $\Bbb{F}$, and let $T:V \to W$ be a linear invertible map. By $T^*$, I assume you mean the map $T^* : W^* \to V^*$ defined by $T^*(f) = f \circ T$ (the dual map/transpose map/pull-back or whatever you want to call it).
So, we need to show that
\begin{align}
T^* \circ \left(T^{-1} \right)^* = \text{id}_{V^*} \quad \text{and} \quad \left(T^{-1}\right)^* \circ T^* = \text{id}_{W^*}
\end{align}
To prove the first assertion, let $f \in V^*$ by arbitrary. Then,
\begin{align}
\left[T^* \circ (T^{-1})^* \right](f) &:=T^* \left( \left(T^{-1} \right)^*[f] \right) \\
&:= T^* \left( f \circ T^{-1} \right) \\
&:= \left(f \circ T^{-1}\right) \circ T \\
&= f \\
&:= \text{id}_{V^*}(f)
\end{align}
Since this is true for every $f \in V^*$, we have just proven that $T^* \circ \left(T^{-1} \right)^* = \text{id}_{V^*}.$ 
In the above string of equalities, $:=$ means "true by definition"; for example the first is true by definition of composition, the 2nd is by definition of $(T^{-1})^*$ etc. I'll leave it to you to prove the other equality. 
