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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?

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    $\begingroup$ 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. $\endgroup$ May 28, 2019 at 14:18
  • $\begingroup$ @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? $\endgroup$
    – gen
    May 28, 2019 at 14:57
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    $\begingroup$ 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). $\endgroup$ May 28, 2019 at 15:13
  • $\begingroup$ @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. $\endgroup$
    – Arnaud D.
    May 28, 2019 at 15:45

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The answer is no for both questions.

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.

You can also find plenty of counterexamples to your second question in this MO thread.

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  • $\begingroup$ 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. $\endgroup$
    – bbarker
    Dec 31, 2019 at 14:59
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    $\begingroup$ @bbarker It need not mean much. See the counterexample I've added. $\endgroup$
    – Arnaud D.
    Jan 1, 2020 at 18:24

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