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Given two categories $\mathcal{C}$ and $\mathcal{D}$, a covariant functor $T:\mathcal{C}\to\mathcal{D}$ consists of an assignment for each object $X$ of $\mathcal{C}$ an object $T(X)$ of $\mathcal{D}$ (i.e. a “function” $T:{\rm Ob}(\mathcal{C})\to{\rm Ob}(\mathcal{D})$) together with an assignment for every morphism $f\in{\rm Hom}_{{\mathcal{C}}}(A,B)$, to a morphism $T(f)\in{\rm Hom}_{{\mathcal{D}}}(T(A),T(B))$, such that:

$T(1_{A})=1_{{T(A)}}$ where $1_{X}$ denotes the identity morphism on the object $X$ (in the respective category).

$T(g\circ f)=T(g)\circ T(f)$, whenever the composition $g\circ f$ is defined.
A contravariant functor $T:\mathcal{C}\to\mathcal{D}$ is just a covariant functor $T:\mathcal{C}^{{\rm op}}\to\mathcal{D}$ from the opposite category. In other words, the assignment reverses the direction of maps. If $f\in{\rm Hom}_{{\mathcal{C}}}(A,B)$, then $T(f)\in{\rm Hom}_{{\mathcal{D}}}(T(B),T(A))$ and $T(g\circ f)=T(f)\circ T(g)$ whenever the composition is defined (the domain of $g$ is the same as the codomain of $f$).
Given a category $\mathcal{C}$ and an object $X$ we always have the functor $T:\mathcal{C}\to{\bf Sets}$ to the category of sets defined on objects by $T(A)={\rm Hom}(X,A)$. If $f:A\to B$ is a morphism of $\mathcal{C}$, then we define $T(f):{\rm Hom}(X,A)\to{\rm Hom}(X,B)$ by $g\mapsto f\circ g$. This is a covariant functor, denoted by ${\rm Hom}(X,)$.
Similarly, one can define a contravariant functor ${\rm Hom}(,X):\mathcal{C}\to{\bf Sets}$.
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