# Generator of a category

I have a question on the definition of a generator of a category $C$. I have made a stupid mistake in the original formulation of the question. I beg your pardon for this.

Let $G$ be a generator of $C$. Let $f:X\to Y$ be a $C$-morphism. Does $f_*:Hom(G,X)\cong Hom(G,Y)$ imply that $f$ is an isomorphism? If yes, is this property for all $C$-morphisms $f$ equivalent to $G$ being a generator of $C$?

Original question: Let $G$ be a generator of $C$. Does $Hom(G,X)\cong Hom(G,Y)$ imply $X\cong Y$? If yes, is this property for all objects $X$ and $Y$ equivalent to $G$ being a generator of $C$?

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What structure are you attaching to the hom-sets? Just the set structure? If you only have the set structure, then any two hom-sets with the same cardinality will be isomorphic, and this property will fail. (Take for example, $C$ = category of abelian groups, $G = Z$, $X = Z, Y=Z \times Z.$) Or do you want a natural transformation of functors? – only Aug 28 '12 at 21:44
A strong generator has the property you want. – Zhen Lin Aug 29 '12 at 0:40
Thanks. A strong generator is a generator but the opposite is not always true, I suppose. Is this opposite true if I require some conditions on the category $C$? I think, that it is true in Sets for example. – Daniel Dreiberg Aug 29 '12 at 6:18

The answer is no. Consider a poset with one least element $0$ and two incomparable elements, $A$ and $B$ both greater than $0$. Then $0$ is a generator (vacuously, because there is never more than one morphism between two objects), and $\hom(0,A)\cong \hom(0,B)$, as they are both sets with one element. However, since $A$ and $B$ are incomparable, they are not isomorphic.
The problem is that there is no structure to the isomorphism $\hom(G,X)\cong \hom(G,Y)$. The isomorphism isn't natural, it isn't induced by a particular map, it's just there. There may be a way to add structure to the problem to get something Yoneda-esque, but as currently formulated, just being a generator is not enough to give what you want.