Why can't covariant? I mean, a functor is made of two data. Part one: object$\rightarrow$object . And Part two: morphism$\rightarrow$morphism . Why can't there exist a functor whose part one is like V$\rightarrow$V*(V* means the dual space) and part two is covariant?why must contravariant? I cannot give a construction of a covariant functor. But it doesn't mean there doesn't exist one. Can we use any method to prove such a functor doesn't exist?
If you still cannot fully understand my question, please look at this picture. An example
My question is for the last sentence "I'm not sure if there's a way to do this---the arrows simply don't line up". This is not a math proof. And I want a math proof of this fact.
Or if I am wrong. There actually exist such a functor. Then can you give a construction or proof of the existence.
Thank you!
Obviously there exists a covariant functor from $Vect_{\mathbb{F}}$ to itself which takes $V$ to $V^*$. Call this functor $F$. Apply a super-strong version of the axiom of choice to pick a basis $(e^V_i)_{i\in I_V}$ for every $V$ and for given linear $f:V\to W$ given by $$ e_i^V\mapsto \sum_{j\in I_W} f_j e^W_j, $$ we simply define $F(f)((e_i^V)^*)=\sum_{j\in I_W} f_j (e_j^W)^*$, where $v^*$ of course denotes the dual-basis vector to $v$ given the chosen bases. Since linear maps are uniquely given by how they act on basis elements, this does give a well-defined map. It's also trivial to check that $F$ maps compositions to compositions.
Now, of course, this functor is incredibly silly and not very interesting. But it does exist.