$\newcommand{\im}{\operatorname{im}}$Theorem: the dual functor on the category of vector spaces over $K$ is exact.
Proof:
Let $U \overset f \longrightarrow V \overset g \longrightarrow W$ be an exact sequence, i.e. $\im f = \ker g$.
Let $f^* : v^* \mapsto v^* \circ f$ and $g^* : w^* \mapsto w^* \circ g$.
To prove that $W^* \overset {g^*} \longrightarrow V^* \overset {f^*} \longrightarrow U^*$ is exact, i.e. $\im g^* = \ker f^*$.
Let $v^* \in V^*$ such that $v^* \in \im g^*$, i.e. there is $w^* \in W^*$ such that $w^* \circ g = v^*$. Then, for every $v \in \ker g$, $v^*(v) = w^*(g(v)) = 0$…
Proof:
Let $U \overset f \longrightarrow V \overset g \longrightarrow W$ be an exact sequence, i.e. $\im f = \ker g$.
Let $f^* : v^* \mapsto v^* \circ f$ and $g^* : w^* \mapsto w^* \circ g$.
To prove that $W^* \overset {g^*} \longrightarrow V^* \overset {f^*} \longrightarrow U^*$ is exact, i.e. $\im g^* = \ker f^*$.
Let $v^* \in V^*$ such that $v^* \in \im g^*$, i.e. there is $w^* \in W^*$ such that $w^* \circ g = v^*$. Then, for every $v \in \ker g$, $v^*(v) = w^*(g(v)) = 0$…
There are a number of notions of infinity in mathematics that are respectable. One of the first is 'the point at infinity' to the line or plane; but one can argue that this is a spurious infinity as in another perspective this simply closes up the line to make a circle; and the plane to make the ...
