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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 $hom(-,a)$ 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.

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A natural transformation is not a map $\textrm{Hom}(x, a) \to F x$, it is a family of such maps satisfying certain equations. –  Zhen Lin Apr 14 '13 at 0:57
    
In other words, you should recapitulate the definitions. –  Martin Brandenburg Apr 14 '13 at 0:58
    
I updated them. Not sure if it's any better though. –  user21154 Apr 14 '13 at 1:33
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Maybe you find Tom Leinster's The Yoneda Lemma: What's It All About? illuminating. –  Martin Apr 14 '13 at 9:23
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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 $F$.

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Thank-you very much. Yeah, I see what you mean. I guess I'm not clear then on how there can even exit more than one natural transformation between two functors ($Hom(-,a)$ and $F a$) because they are indexed by elements in C the natural transformation sends Hom(c,a) to F c. –  user21154 Apr 14 '13 at 12:33
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