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I have a cocomplete abelian category $\mathcal C$ and two objects $X$, $Y$ in $\mathcal C$. Further, $\mathcal C$ has a projective generator $P$. I have an isomorphism

$$ \mathcal C(P,X) \cong \mathcal C(P,Y)$$

It feels like this should be enough to show that $X\cong Y$ in $\mathcal C$. Because any object $Z$ in $\mathcal C$ can be written as a cokernel $P^J\longrightarrow P^I\longrightarrow Z\longrightarrow 0$ and so we have $\mathcal C(Z,X)\cong \mathcal C(Z,Y)$. It feels like the isomorphisms $\mathcal C(Z,X)\cong \mathcal C(Z,Y)$ are natural in $Z$ and do not depend on the choice of projective resolutions.

But I cannot be sure. Is this true? Any reference or pointers would be appreciated.

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  • $\begingroup$ Your claim is false. You can find a counterexample where $\mathcal{C}$ is the category of real vector spaces. $\endgroup$ – Zhen Lin Jun 15 at 4:45
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    $\begingroup$ Isomorphism of $\mathcal{C} (P,X)$ with $\mathcal{C} (P,Y)$ as what? If you consider them as $\text{End}(P)^{\text{op}}$-modules, then yes, this implies the isomorphism of objects. $\endgroup$ – Sasha Jun 15 at 8:10
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It is very easy to specify an answer if $P$ is a compact projective generator (you didn't write whether it is compact). Then, we have an equivalence of categories between $\mathcal{C}$ and the category of right modules over $\text{End} (P)$, given by sending $X$ to $\mathcal{C} (P , X)$. So, of course, if objects become isomorphic under an equivalence of categories, they were isomorphic. For that answer, one also needs to understand your isomorphism as compatible with the endomorphisms of $P$, i.e. a $\text{End}(P)$-module isomorphism.

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