Understanding the Yoneda lemma I'm having difficulty understanding the Yoneda lemma. In particular, the proof isn't that obvious to me. Please, could someone explain to me the error of my current understanding..
The Yoneda lemma says that if $F$ is a functor from $C \rightarrow Set$, then $F a$ is isomorphic to the natural transformations from ${\rm Hom}(-,a)$ to $F$. Now, the part that gets me is if I say $F$ is a functor which sends objects of $C$ to the same single-element set and morphisms to the identity function, how can this bijection exist. There is one element in $F a$, but multiple natural transformations (e.g. $C(-,a)$ to $F$, and $C(-,b)$ to $F$).
Please someone help me.
 A: The main error seems to be in the last parenthetical example of your question, where $b$ appeared out of nowhere.  Yoneda's Lemma says that $Fa$ is isomorphic to the set of natural transformations from Hom$(-,a)$ to $F$, and similarly $Fb$ is isomorphic to the set of natural transformations from Hom$(-,b)$ to $F$, but it says nothing that connects $Fa$ with natural transformations from Hom$(-,b)$ to $F$.  
A: If $\forall a \in C,\ Fa \cong \{*\}$ ($Fa$ is a set with a single element), then all $a \in C$ are mapped to the terminal object of the Set category.
Being a terminal object means that, $\forall X \in Set$, there is one single function $X \to \{*\}$ (up to isomorphism).
Considering that a natural transformation
$$\alpha:{\rm Hom}(-,a) \to F$$
is a set of functions of type: 
$$\alpha_c:{\rm Hom}(c,a) \to Fc,\quad \forall c \in C$$
and considering the above definition of $F$, then the $\{\dots,\alpha_c,\dots\}$ are functions to the terminal object:
$$\alpha_c:{\rm Hom}(c,a) \to \{*\},\quad \forall c \in C$$
Finally, there is one single set of functions to define $\alpha$ given the definition of the terminal object: the set of unique functions that send the objects of Set to the terminal object (up to isomorphism).
