Prove that a morphism $\alpha$ of $Fun(\mathcal{A},\mathcal{B})$ is an isomorphism iff each component $\alpha_A$, is an isomorphism in $\mathcal{B}$ I'm a computational engineer starting with a course of Introduction to Category Theory, and perhaps is extremely basic what I'm asking but I'm trying to learn how to make proofs in category theory working with the very basic concepts, but I just can't figure things out right now. Can you help me?
Thank you!

Consider a small category $\mathcal{A}$ and the functor category $Fun(\mathcal{A},\mathcal{B})$. Prove that a morphism $\alpha$ of $Fun(\mathcal{A},\mathcal{B})$ (a natural transformation) is an isomorphism if, and only if, each component $\alpha_A$, $A\in \mathcal{A}$, is an isomorphism in $\mathcal{B}$.

 A: If $F,G:\mathcal A\rightarrow\mathcal B$ are functors and $\alpha:F\stackrel{\bullet}{\rightarrow}G$ is a natural transformation
such that $\alpha_{A}:F\left(A\right)\rightarrow G\left(A\right)$
is invertible for each object $A\in\mathcal A$ then it can be shown that the inverses $\alpha_{A}^{-1}:G\left(A\right)\rightarrow F\left(A\right)$
are the components of a natural transformation $G\stackrel{\bullet}{\rightarrow}F$. 
We have $\alpha_{A}\circ\alpha_{A}^{-1}=\text{id}_{G\left(A\right)}$
and $\alpha_{A}^{-1}\circ\alpha_{A}=\text{id}_{F\left(A\right)}$
for each object $A\in\mathcal{A}$, so this natural transformation
appears to be the inverse of $\alpha$.
If conversely $\alpha:F\stackrel{\bullet}{\rightarrow}G$ is a natural
isomorphism and $\beta:G\stackrel{\bullet}{\rightarrow}F$ denotes
its inverse then $\alpha\circ\beta=\text{id}_{G}$ and $\beta\circ\alpha=\text{id}_{F}$
wich comes the same as $\alpha_{A}\circ\beta_{A}=\text{id}_{G\left(A\right)}$
and $\beta_{A}\circ\alpha_{A}=\text{id}_{F\left(A\right)}$ for each
object $A$ of $\mathcal{A}$. This shows that the $\alpha_{A}$ are
all invertible.
