>**Puzzle 283**

>Guess what the cap and cup

>$$ \cap_V \colon k \to V \otimes V^\ast, \qquad \cup_V \colon V^\ast \otimes V \to k $$

>are for a finite-dimensional vector space \\(V\\), and check your guess by proving the snake equations.

\\(\cap_V \colon k \to V \otimes V^\ast \\) is the function that takes the inverse of a matrix of a vector space which sends it into its dual space. \\(\cup_V \colon V^\ast \otimes V \to k\\) is the function that takes the inverse of a dual matrix which sends it back into the vector space.

The snake equations are composing the two inverse functions and therefore, come back full circle back to what you start with.

More precisely, Let \\(\mathbf{V}\\) be a matrix in a vector space and \\(\mathbf{V^\ast}\\) be matrix in its dual space. Then:

\[\mathbf{V^\ast} \cdot \mathbf{V} = \mathbf{I} = \mathbf{V} \cdot \mathbf{V^\ast}\]

\[\mathbf{V^\ast} = \mathbf{V}^{-1}\]

The snake equations would yield:

\[\mathbf{V} = \mathbf{I} \cdot \mathbf{V} = (\mathbf{V}\cdot \mathbf{V^\ast}) \cdot \mathbf{V} = \mathbf{V}\cdot (\mathbf{V^\ast} \cdot \mathbf{V}) = \mathbf{V} \cdot \mathbf{I} = \mathbf{V} \]

\[\mathbf{V^\ast} = \mathbf{V^\ast} \cdot \mathbf{I} = \mathbf{V^\ast}\cdot (\mathbf{V} \cdot \mathbf{V^\ast}) = (\mathbf{V^\ast}\cdot \mathbf{V}) \cdot \mathbf{V^\ast} = \mathbf{I} \cdot \mathbf{V^\ast} = \mathbf{V^\ast} \]

An alternative way to show this is:

\[(\mathbf{V}^{-1})^{-1} = \mathbf{V}\]

\[(\mathbf{V^\ast}^{-1})^{-1} = \mathbf{V^\ast}\]

>Guess what the cap and cup

>$$ \cap_V \colon k \to V \otimes V^\ast, \qquad \cup_V \colon V^\ast \otimes V \to k $$

>are for a finite-dimensional vector space \\(V\\), and check your guess by proving the snake equations.

\\(\cap_V \colon k \to V \otimes V^\ast \\) is the function that takes the inverse of a matrix of a vector space which sends it into its dual space. \\(\cup_V \colon V^\ast \otimes V \to k\\) is the function that takes the inverse of a dual matrix which sends it back into the vector space.

The snake equations are composing the two inverse functions and therefore, come back full circle back to what you start with.

More precisely, Let \\(\mathbf{V}\\) be a matrix in a vector space and \\(\mathbf{V^\ast}\\) be matrix in its dual space. Then:

\[\mathbf{V^\ast} \cdot \mathbf{V} = \mathbf{I} = \mathbf{V} \cdot \mathbf{V^\ast}\]

\[\mathbf{V^\ast} = \mathbf{V}^{-1}\]

The snake equations would yield:

\[\mathbf{V} = \mathbf{I} \cdot \mathbf{V} = (\mathbf{V}\cdot \mathbf{V^\ast}) \cdot \mathbf{V} = \mathbf{V}\cdot (\mathbf{V^\ast} \cdot \mathbf{V}) = \mathbf{V} \cdot \mathbf{I} = \mathbf{V} \]

\[\mathbf{V^\ast} = \mathbf{V^\ast} \cdot \mathbf{I} = \mathbf{V^\ast}\cdot (\mathbf{V} \cdot \mathbf{V^\ast}) = (\mathbf{V^\ast}\cdot \mathbf{V}) \cdot \mathbf{V^\ast} = \mathbf{I} \cdot \mathbf{V^\ast} = \mathbf{V^\ast} \]

An alternative way to show this is:

\[(\mathbf{V}^{-1})^{-1} = \mathbf{V}\]

\[(\mathbf{V^\ast}^{-1})^{-1} = \mathbf{V^\ast}\]