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Objects that are isomorphic to a terminal object are terminal themselves, and so the isomorphism turns to be unique. I am looking for an example of a limit or colimit (the vertex of the universal cone will be denoted $C$), such that there exists an object $C'$ (which maybe could be $C$ itself) which is isomorphic to $C$ but not uniquely so. I have tried to look this up but could not find any reference where this comes up.

Thank you.

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  • $\begingroup$ I don't get your question: elements defined by a unique property all enjoy the same kind of strong unicity property: en.wikipedia.org/wiki/… so that also holds for (co)limits. $\endgroup$ – PseudoNeo Aug 23 '16 at 14:33
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    $\begingroup$ I am sorry for the poor wording. Here is one setting: say you have an isomorphism of rings $C \to A \times B$. You can equip $C$ with projections, turning it into "a product" of $A$ and $B$, but (I guess) that does not mean there was only one isomorphism $C \to A \times B$ to begin with. $\endgroup$ – Cotton Aug 23 '16 at 14:39
  • $\begingroup$ Maybe I am misunderstanding your comment, but I see it as follows: there may indeed be more than one isomorphism $\phi:C\to A\times B$ but each gives rise to a limit cone $(C,\pi_i\circ \phi)$. The uniqueness lies in the fact that, for a given pair $(f_1,f_2)$ there is only one$\phi$ such that $f_i=\pi_i\circ \phi$ $\endgroup$ – Matematleta Aug 23 '16 at 15:03
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    $\begingroup$ Yes, we are on the same page. $\endgroup$ – Cotton Aug 23 '16 at 15:47
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This will happen more or less whenever you have an object with a non-trivial automorphism.

For example, in most abelian groups the automorphism $a\mapsto-a$ is non-trivial. If $G$ is such an abelian group (say $\mathbb{Z}$ under addition if you want a concrete example) then the product $G\times G$ has an automorphism which is different from the identity. (Actually it has at least three, because you can negate either component or both.)

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