Is there a reason why we call $\mathrm{Hom}(A, -)$ covariant and $\mathrm{Hom}(-,B)$ contravariant? I have trouble remembering what it means to apply the covariant or contravariant Hom functor.
For example, when I see $\mathrm{Hom}(A, -)$, I always forget if I am going to reverse arrows or not.
For example: I used to have trouble remembering what a pushout and pullback was until someone pointed out that pushout diagrams have arrows going forward from the set that is going to dictate the gluing. Now I will never forget what a pushout diagram is.
Does anyone have any mnemonics to help me remember which is the covariant and contravariant hom functor?
 A: Covariant means having the same "variance", or in other words going in the right direction. Here it means that if you apply $\hom(A,-)$ to an arrow $f:C\to D$, you get an arrow in the same direction $\hom(A,C)\to \hom(A,D)$; hence covariant.
If you apply $\hom(-,B)$ to the same arrow you get an arrow $\hom(D,B)\to \hom(C,B)$ so in the opposite direction, hence contravariant. I guess it's easier if your mother tongue is a latin language, because in those "contra" (or derivatives thereof) often indicates some notion of "opposite" or "against".
A: I am not sure what precisely you are asking, so I shall provide a blanket answer.
Notation of Hom-sets
Notationally, $Hom(A, B)$ has function from $A$ to $B$. I think of $Hom(A, B) \equiv \{ A \rightarrow B \}$. Given this notation, it's somewhat natural to see that $Hom(A, -) = \{ A \rightarrow - \}$, so we have applied $A$ to $Hom$, and we are waiting for the other argument. So, the set we are coming from is fixed ($A$). $Hom(A, -) \equiv \text{maps out of $A$}$.
Similarly, $Hom(-, B) \simeq (- \rightarrow B)$ has $B$ fixed, and the domain unfixed. So, $Hom(-, B)$ is maps into B.
Now, the co/contravariance is as follows:
Contravariance
Imagine we have a map $f: X \rightarrow A$, and $Hom(A, -)$ with us. Then, we can create $Hom(X, -) \equiv \{  h \circ f \vert h \in  Hom(A, -) \}$. That is, we get a mapping:
$$(f: X \rightarrow A) \rightarrow (h: Hom(A, -) \rightarrow Hom(X, -))$$
This mapping reverses the direction of the sets. We have $X \rightarrow A$, which flipped to become $(Hom(A, -) \rightarrow Hom(X, -))$. Hence, this is contravariant (as in, contrast / opposite).
Covariance
Once again, we have a map $f: X \rightarrow A$, and $Hom(-, X)$. We can convert this to $Hom(-, A) \equiv \{ f \circ h ~ \vert ~ h \in Hom(-, X) \}$. Hence, this gives us a function:
$$(f: X \rightarrow A) \rightarrow (h: Hom(-, X) \rightarrow Hom(-, A))$$
