# Isomorphism of hom sets implies objects are isomorphic

Suppose we have

$$Hom(x_0, y) \simeq Hom(x_1, y)\ \forall y \in \mathcal{C}$$

Does it follow that $$x_0 \simeq x_1$$ in $$\mathcal{C}$$?

Now suppose that $$F, G: \mathcal{C} \rightarrow \mathcal{D}$$ are two functors such that $$Fx \simeq Gx\ \forall x\in \mathcal{C}$$. Does it follow that any natural transformation $$\epsilon: F \Rightarrow G$$ is a natural isomorphism?

• Your question already has an excellent answer, but it may be worth pointing out that if you require $\operatorname{Hom}(x_0, y) \simeq \operatorname{Hom}(x_1, y)$ to be natural in $y$, then you can conclude that $x_0$ and $x_1$ are isomorphic. May 28, 2019 at 14:18
• @MarkKamsma thank you for pointing this out, this is might just be what I am after here... Could you please explain what naturality means in this context?
– gen
May 28, 2019 at 14:57
• Given an arrow $f: y \to y'$ we can consider $\operatorname{Hom}(x_0, f): \operatorname{Hom}(x_0, y) \to \operatorname{Hom}(x_0, y')$ by composing with $f$, and similarly with $x_1$. Then saying that the bijection is natural in $y$ means that for every $f: y \to y'$ the relevant square commutes. This is a bit hard to draw in a comment, but it means that $\operatorname{Hom}(x_0, y) \simeq \operatorname{Hom}(x_1, y) \to \operatorname{Hom}(x_1, y')$ is the same as $\operatorname{Hom}(x_0, y) \to \operatorname{Hom}(x_0, y') \simeq \operatorname{Hom}(x_1, y')$ (draw this yourself). May 28, 2019 at 15:13
• @gen What Mark said in his first comment is the subject of this question. I should perhaps add for completeness that this is why I said that $F$ and $G$ are not isomorphic if $x_0$ and $x_1$ are not. May 28, 2019 at 15:45

For the first one, define a category $$\mathcal{C}$$ with two objects $$x_0,x_1$$, the set of morphisms $$x\to y$$ is $$\{0,1,\dots\}$$ if $$x=y$$ and $$\{1,2,\dots\}$$ otherwise, and composition is just addition of natural numbers. Then for all $$y$$ there is a bijection between $$\operatorname{Hom}(x_0,y)$$ and $$\operatorname{Hom}(x_1,y)$$ (since both are infinite and countable), but there is no isomorphism $$x_0\to x_1$$.
For a less artificial counterexampe, take the category of finite-dimensional vector spaces over $$\mathbb{R}$$; then $$\operatorname{Hom}(x,y)$$ is in bijection with $$\mathbb{R}^{\dim(y)\times \dim(x)}$$, so it has the same cardinality as $$\mathbb{R}$$ as long as neither $$x$$ nor $$y$$ is the zero vector space. In particular, for all non-zero vector spaces $$x_0,x_1$$, there is a bijection between $$\operatorname{Hom}(x_0,y)$$ and $$\operatorname{Hom}(x_1,y)$$ for all $$y$$.
For the second question, take any counterexample to the first one, and take $$\mathcal{D}=\mathbf{Set}$$ and $$F,G$$ the functors represented by $$x_0$$ and $$x_1$$ respectively. Then there exist morphisms $$x_0\to x_1$$ and $$x_1\to x_0$$, and thus natural transformations $$G\Rightarrow F$$ and $$F\Rightarrow G$$, but $$F$$ and $$G$$ are not isomorphic since $$x_0$$ and $$x_1$$ aren't. Note that even if $$F$$ and $$G$$ were isomorphic, there could still be natural transformations between them that are not isomorphisms.
• In the first case, is there some notion of equivalence or special relationship between $x_0$ and $x_1$? It seems like this is a fairly strong statement of similarity between $x_0$ and $x_1$, at least within the Category in question. Dec 31, 2019 at 14:59