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If an object $X$ has a non-trivial automorphism $g$, for any isomorphism $f$ with an object $Y$ there is another isomorphism $f \circ g$ between $X$ and $Y$, so there is not a unique isomorphism between $X$ and $Y$.

Does the second "unique" in "unique up to unique isomorphism" mean nothing else than that the object in question has no automorphism than the identity?

If the answer is positive: Why is this second - and somehow independent - fact so strongly interwoven (terminologically) with the fact of being unique up to isomorphism?

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It is vary rare that the expression is used in such a context: it is usually used in a context such as the statement of the uniqueness up to unique isomprphism of a product in a category, in which "up to isomorphism" involves more structure than the object itself. –  Mariano Suárez-Alvarez Jul 3 '11 at 19:39
    
In particular, it is true, in the usual sense of the word, that a product of two objects in a category is, when it exists, unique up to unique isomorphism, but this statement makes absolutely no assertion about the existence of automorphisms of the product object. Indeed, the product of two objects usually has lots of automorphisms! –  Mariano Suárez-Alvarez Jul 3 '11 at 19:42
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I'm not really sure how this is [set-theory], could you explain? –  Asaf Karagila Jul 3 '11 at 21:16
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I would say that the locution "unique up to unique isomorphism" is slightly jargony. Of course it does not mean unique in the literal sense: it means there is a unique isomorphism such that X holds, where $X$ is generally some universal mapping property. When I learned this material, the turn of phrase I heard was "unique up to canonical isomorphism", which I think is more accurate. –  Pete L. Clark Jul 4 '11 at 1:40

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up vote 17 down vote accepted

When people say, for example, that the product $X \times Y$ of two objects $X, Y$ is unique up to unique isomorphism, that doesn't mean that $X \times Y$, as an object of the category, has no non-trivial automorphisms; it's easy to find examples where this is blatantly false. It means that if you have two objects $A, B$ which are both products of $X$ and $Y$ in the sense that they come with distinguished projection maps to $X$ and $Y$ satisfying the universal property, then there is a unique isomorphism $A \to B$ compatible with the projection maps. The projection maps are part of the data that defines a product, and in particular it is possible for the same object to be a product of $A$ and $B$ in two different ways (in the sense that the projection maps are different): those different ways are then related by an automorphism of the object.

Another way to say this is to say that a product is a terminal object in a certain category of cones, and as an object of this category, it follows that the product has no non-trivial automorphisms because, for any terminal object $1$, there is a unique map $1 \to 1$, which must be the identity.

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