Let C be a locally small category (i.e. a category for which hom-classes are actually sets and not proper classes).
For all objects A and B in C we define two functors to the category of sets as follows:
Hom$(-,B):C\rightarrow$ Set
This is a contravariant functor given by:
- Hom$(-,B)$ maps each object $X$ in $C$ to the set of morphisms, Hom$(X,B)$
- Hom$(-,B)$ maps each morphism $h:X\rightarrow Y$ to the function: Hom$(h,B)$:m Hom$(Y,B)\rightarrow $Hom$(X,B)$ given by $g\mapsto g\circ h$ for each $g$ in Hom$(Y,B)$.
But I'm not sure how to intemperate the Hom$(h,B)$ in thus case. $B$ is a functor which is also a morphism, a single object. $h$ is a morphism but not a functor, and it maps the object $X$ in $C$ to some unknown range. I don't understand why Hom$(h,B)$ is possible since, for Hom$(M,N)$, the $M$ and $N$ has to be set. (See: What does Hom(M,N) mean? Atiyah Macdonald proposition 2.9)
What does $Hom(h, B)$ mean in the contravariant functor?
Related:
Here $Hom(h,B)$ is just a notation, we have that $Hom(-,B)$ is a functor, and so must send objects to objects and morphisms to morphisms (it is like 2 maps in one notation). Thus if $h:X\rightarrow Y$ is a morphism in the category $C$, where $X,Y$ and $B$ live as objects, then $Hom(h,B)$ is a map from $Hom(Y,B)$ to $Hom(X,B)$ (here the $Y$ comes first because the functor is contravariant) which sends a map $f:Y\rightarrow B$ to the map $Hom(h,B)(f)=f\circ h:X\rightarrow Y$.