# What does $Hom(h, B)$ mean in the contravariant functor?

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:

The definition of contravariant hom functor

On the definition of the contravariant hom functor

• I'm quoting the same wikipedia page you linked at the top of your post: $\text{Hom}(–, B)$ maps each morphism $h : X \to Y$ to the function $\text{Hom}(h, B) : \text{Hom}(Y, B) \to \text{Hom}(X, B)$ given by $g \mapsto g \circ h$ for each $g \in \text{Hom}(Y, B)$ What's your confusion? Commented Mar 26 at 23:50
• "$B$ is a functor which is also a morphism" It is neither. $B$ is just an object of $C$. Also the range of $h$ is irrelevant: Only its domain and codomain are important. Commented Mar 26 at 23:55
• @BenSteffan Thank you. But If $B$ is an object in $C$ and $Y$ an X an object in $C$, what does Hom$(X,B)$ mean? (Quote: math.stackexchange.com/a/1806668/603316 "Hom(M,N) refers to the set of A-module homomorphisms from M to N.") I'm think about the set notation, $X=\{\{\}\}$ and $B=\{\{\{\}\}\}$ the only morphism is to send $\{\}$ to $\{\{\}\}$. I don't understand how Hom(Y,B) and Hom(X,B) were possible Commented Mar 27 at 0:20
• @ShoutOutAndCalculate As quoted by you, $\operatorname{Hom}(X, B)$ is the set of all morphisms $f\colon X \to B$. Thus, in your case, $\operatorname{Hom}(X, B) = \{g\}$ where $g\colon X \to B$ is the unique map. Commented Mar 27 at 13:14
• @ShoutOutAndCalculate Not strange at all. And no, it should not be $\operatorname{Hom}(\{X\}, \{B\})$. $\operatorname{Hom}({{-}}, {{-}})$ takes two objects as input, such as $X$ and $B$. The notation $\{X\}$ does not even make sense in a general category. Commented Mar 27 at 14:29

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 $$Hom(h,B)(f)=f\circ h:X\rightarrow Y$$.