I've recently started to study category through Leinster's Basic Category Theory.
In example 1.2.12 it's said that $\textbf{Hom}(V,W)$ is a vector space.
Then,
Now fix a vector space $W$. Any linear map f : $V \rightarrow V′$ induces a linear map
$f^*:\textbf{Hom}(V',W)\rightarrow \textbf{Hom}(V,W)$ such that for $q\in \textbf{Hom}(V',W)$ we have $f^*(q)= q f$
He concludes by saying that
This defines a functor $\textbf{Hom}(-,W):\textbf{Vect}^{op}_k\rightarrow \textbf{Vect}_k$
My problem is that I don't understand why is the op necessary there. Why wouldn't it work without it?
Best Answer
The op is necessary there because one would like to work with covariant functors $F: \mathcal{C} \rightarrow \mathcal{D}$ between categories which preserve the order of composition: i.e., given a map $f: C_1 \rightarrow C_2$ in $\mathcal{C}$ then we would like $F(f): F(C_1) \rightarrow F(C_2)$ in $\mathcal{D}$. The trouble with this is that in this case, $Hom(-,W)$ to a fixed vector space $W$ is contravariant in the first argument, as you explain above.
We need the op in order to have a covariant functor, in this case from $Vect^{op}_k \rightarrow Vect_k$.