Via Wikipedia I see that $\operatorname{Hom}_C(A,-): C \rightarrow \textbf{Set}$ a covariant functor which maps each object $X$ in $C$ to the set of morphisms $\operatorname{Hom}_C(A,X)$. I am trying to understand this better (the point and applications) as to me as it seems that all one is doing is mapping $X$ to all of the possible morphisms $A \rightrightarrows X$, so $X$ is going to some bundle of morphisms in which it is an element of (it is being mapped to a set of maps that map to itself).
What new information does this association give?
Digging deeper say we have elements $X,Y$,and $Z \in C$. I can see that we have 3 bundles $\operatorname{Hom}_C(A,X),\operatorname{Hom}_C(A,Y)$, and $\operatorname{Hom}_C(A,Z)$ and that $\operatorname{Hom}_C(A,-)$ is the set of all possible bundles of morphisms from $A$ to any object in $C$, but why have each object i.e. $X,Y,Z$ also map to these bundles?
Thanks,
Brian
Echoing what Jakob said, no 'new' information is gained, but rather it provides us a tool with which we can express that information in an easily manipulable way.
For example, a category theoretic fact that makes it a little more clear why such a form would more easily manipulable: $$\hom(A,\lim F)\cong \lim \hom(A,F(-))$$ and similarly $$\hom(\mathrm{colim} \,F,B)\cong \lim \,\hom(F(-),B)$$ where each of these bijections are natural. This here makes proving some statements extremely simple (for example, the adjunction of the geometric realization and singular simplicial complex follows rather easily by using these tools).
Additionally, adjunctions show up often in category theory; for example, in Cartesian Closed Categories (CCCs), there is an adjunction between the product and the exponential. When proving things about objects for which such adjunctions exist, being able to translate into the formulation of these adjunctions gives an indispensable tool.
For example, we will prove that for CCC we have $A^{B\times C}\cong (A^B)^C$ (assuming the Yoneda Lemma, which Jakob mentioned):
\begin{align} \hom(X,A^{B\times C}) & \cong \hom(X\times (B\times C),A) \\ &\cong \hom(X\times (C\times B),A) \\ & \cong \hom((X\times C)\times B,A) \\ & \cong \hom(X\times C,A^B) \\ & \cong \hom(X,(A^B)^C) \end{align} From which by the Yoneda Lemma we conclude that $A^{B\times C}\cong (A^B)^C$.
Similarly, for a CCC with coproducts (I'll denote it by $\oplus$) and products distributing over these, we have $A^{B\oplus C}\cong A^B\times A^C$:
\begin{align} \hom(X,A^{B\oplus C}) & \cong \hom(X\times (B\oplus C),A) \\ & \cong \hom((X\times B)\oplus (X\times C),A) \\ & \cong \hom(X\times B,A) \times \hom(X\times C,A) \\ & \cong \hom(X,A^B)\times \hom(X,A^C) \\ & \cong \hom(X,A^B\times A^C) \end{align} from which by the Yoneda Lemma we conclude that $A^{B\oplus C}\cong A^B\times A^C$.
(Of course, I made use of other facts in both of these examples; for example the associativity and commutativity of $\times$ in the first and I used the relationships between the $\hom$ functor and limits/colimits in the second.)
Hopefully these examples show you how $\hom$ functors become indispensable when used with the Yoneda Lemma.