I have a couple of doubts about category theory and I would really appreciate help
I began reading Saunders Mac Lane - Categories for Working Mathematician. I have found this excersie For categories $A$, $B$ and $C$ establish natural isomorphisms
a) $(A \times B)^{C} = A^{C} \times B^{C}$
This one seems almost trivial, I think at most you have to use the universal property of the product. So, this is trivial or I'm really lost.
b) $C^{A \times B} \cong (C^{B})^{A}$
This one is a little more tricky and I think I have found a solution but it is not complete.
Given a functor (an object of the category) $H \in (C^{B})^{A}$, we want an element of $C^{A \times B}$.
So, I assign $\Phi(G)=F$ such that $F(a,b)=H(a)(b)$.
So far, so good, it's pretty trivial. Now the arrows are a little more tricky, as I cannot simply name $F(f,g)=H(f)(g)$ as $H(f)(g)$ doesn't makes sense, $H(f)$ is a natural transformation and it only takes values on objects.
So, given arrows $f:a \rightarrow a'$ and $g:b \rightarrow b'$, I assign: $$ F(f,g) = H(a')(g) \circ H(f)(b) $$ Which are arrows on $C$ as $H(a')$ is a functor from $B$ to $C$, so $H(a')(g)$ is an arrow on $C$ and $H(f)$ is a natural transformation between elements on $C$, so $H(f)(b)$ is an arrow in $C$ from $H(a)(b)$ to $H(a')(b)$. So, the given object it is really the image of an arrow from $(a,b)$ to $(a',b')$.
Moreover $$ H(a')(g) \circ H(f)(b)=H(f)(b') \circ H(a)(g) $$ as $H(f)$ is a natural transformation betweem $H(a)$ and $H(a')$.
We want to prove: $$F(f'f,g'g) = F(f',g') \circ F(f,g)$$
Proof
We take arrows $f:a \rightarrow a'$, $f'=a' \rightarrow a''$, $g:b \rightarrow b'$ and $g' : b' \rightarrow b''$.
$$F(f',g') \circ F(f,g) = \left( H(f')(b'') \circ H(a')(g') \right) \circ \left( H(a')(g) \circ H(f)(b) \right) =$$ $$H(f')(b'') \circ \left( H(a')(g') \circ H(a')(g) \right) \circ H(f)(b)=$$ Because $H(a')$ is a functor. $$ = H(f')(b'') \circ H(a')(g'g) \circ H(f)(b) =H(f')(b'') \circ H(f)(b'') \circ H(a)(g'g) $$ Because $H(f)$ it is a natural transformation. I am not sure how to justify the next step, I know that the compose of natural transformations it is a natural transformation, but I am not sure how that implies $H(f')(b'') \circ H(f)(b'') = H(f'f)(b'')$ $$ H(f')(b'') \circ H(f)(b'') \circ H(a)(g'g) = H(f'f) H(b'') \circ H(a)(g'g) = F(f'f,g'g) $$
Now, I need to show that $F(Id_{a},Id_{b})= Id_{F(a,b)} $
$$ F(Id_{a},Id_{b})= H(a)(Id_{b}) \circ H(Id_{a})(b) = Id_{H(a)(b)} \circ H(Id_{a})(b) = Id_{H(a)(b)} \circ Id_{H(a)} (b) $$ Where the second equality is true because $H$ is functor, so $H(Id_{a})=Id_{H(a)}$
After that, $Id_{H(a)} (b) = Id_{H(a)(b)}$ because $Id_{H(a)}$ is an identity natural transformation.
I am not sure if this is a natural isomorphism, moreover the exercise finally ask you: "Compare the second isomorphism with the bijection of Exercise 1". Where exercise 1 ask: For small categories $A$, $B$ and $C$ establish a bijection: $$ \mathbf{Cat} (A \times B, C) \cong \mathbf{Cat} (A,C^{B}) $$ and show it natural in $A$, $B$ and $C$. Hence show that $- \times B : \mathbf{Cat} \rightarrow \mathbf{Cat}$ has right adjoint.
I do not know the difference between exercise 1 and exercise 2)b). I have tried to find in the book the definition of $\mathbf{Cat} (A,B)$ but I had not success. I thought it was the category of the morphisms between $A$ and $B$ and it's natural transformations as arrows, but apparently it is not the same.
Thank you for your time.