A free group $F_S$ over a set $S$ is unique up to unique isomorphism. After reading this answer here I think this means that if $F^\prime_S$ is another free group over $S$ then there exists exactly one isomorphism $\varphi : F_S \to F_S^\prime$ such that $\varphi \circ i = i^\prime$ where $i: S \to F_S$ and $i^\prime : S \to F^\prime_S$ are the inclusions.

Why is uniqueness of the isomorphism interesting at all? Wouldn't it be good enough to know that all free groups over $S$ are isomorphic to define free groups?

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    $\begingroup$ Wouldn't it be enough for what? $\endgroup$ Apr 12, 2012 at 17:20
  • $\begingroup$ @MarianoSuárez-Alvarez Enough to define what a free group is. $\endgroup$ Apr 12, 2012 at 17:20
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    $\begingroup$ But the definition of free groups is completely independent of the fact that they are unique up to unique isomorphism: the latter is a property that one proves after we define what a free group i and even after we prove that they actually exist. $\endgroup$ Apr 12, 2012 at 17:23
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    $\begingroup$ I think he is asking why "unique up to unique isomorphism" is a property worth talking about. I'm not really sure myself(though I've been told it is an important property). Mr. Kent, can you verify if this is what you're asking? $\endgroup$ Apr 12, 2012 at 18:12
  • $\begingroup$ @JohnStalfos Thank you, I'm asking both: what is means and why it's worth talking about. $\endgroup$ Apr 12, 2012 at 19:02

2 Answers 2


One reason you are interested in "unique up to unique isomorphism" is that it makes things more "canonical."

Consider for example the following true fact: every $n$-dimensional vector space over $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$.

However, except for $n=0$, the isomorphism is neither unique nor "canonical": if I go home and do computations based on my favorite isomorphism and I tell you the answer, and you go home and you do your computations based on your favorite isomorphism, and then we come back tomorrow and describe our results, in order to translate one result into "the language of the other" we need to find out what my favorite isomorphism is, what your favorite isomorphism is, and perform the entire translations in order to compare "my" translation into $\mathbb{R}^n$ with yours and find out if we are both correct.

However, when there is a canonical choice of isomorphism, then we can simply both go home and translate (via this unique isomorphism), and then present the results the next day. They must match, because there is essentially only one way to translate.

(Canonicity is very useful; for instance, that is one reason why, still talking about linear algebra, the isomorphism in finite dimension between a vector space and its double dual is so much more important than the isomorphism between a vector space and its dual. The former is canonical, while the latter is not.)

From a point of view of category theory, uniqueness up to unique isomorphism allows for easy functorial properties to be derived, whereas merely isomorphic means a lot more book-keeping is needed, and it may be impossible to state a "coordinate-free" functorial property (where I'm thinking of 'coordinate-free' somewhat fuzzily to refer to not having to invoke specific, arbitrarily chosen isomorphisms to state or prove them).

  • $\begingroup$ Hi @Arturo. I was wondering if you would tell me whether my understanding is correct: If e.g. we're in the category of sets and $A$ is an initial object in Set then it's unique up to isomorphism which means that if $B$ is another initial object then $A$ and $B$ are isomorphic? That is, any two initial objects in a category are isomorphic. Thank you a lot! $\endgroup$ Apr 17, 2012 at 7:42
  • $\begingroup$ I think the answer is yes. $\endgroup$ Apr 17, 2012 at 14:04
  • $\begingroup$ @ClarkKent: Sorry, I didn't see the question. Yes, in any category, two initial objects are isomorphic via a unique isomorphism: if $I$ and $I'$ are both initial, then there is a unique map $f\colon I\to I'$ (because $I$ is initial), a unique map $g\colon I'\to I$ (because $I'$ is initial), and so $f\circ g\colon I'\to I'$ must be the unique map $I'\to I'$ (which is $\mathrm{id}_{I'}$, and $g\circ f\colon I\to I$ must be the unique map $I\to I$ (which is $\mathrm{id}_I$), so $f$ and $g$ are isomorphisms, and they are the unique isomorphisms. In $\mathbf{Sets}$, $\varnothing$ is the initial obj $\endgroup$ Apr 17, 2012 at 15:29

Uniqueness up to unique isomorphism implies that there is no automorphism $f:F_S\to F_S$ which is the identity on the image of $i:S\to F_S$.

  • $\begingroup$ Thank you. I'm trying to understand: reading Wikipedia makes it sound as if essentially unique is a synonym of unique up to unique isomorphism, is this right? $\endgroup$ Apr 12, 2012 at 19:06
  • $\begingroup$ Then maybe "unique up to unique isomorphism" means that for two free groups $(i, F_S), (i^\prime, F_S^\prime)$ there is an isomorphism $\varphi: F_S \to F_S^\prime$ such that $\varphi \circ i = i^\prime$? $\endgroup$ Apr 12, 2012 at 19:07
  • $\begingroup$ @ClarkKent: There is a unique isomorphism with that property; and that follows from the universal property of the free group. $\endgroup$ Apr 12, 2012 at 23:30
  • $\begingroup$ @ArturoMagidin Thank you! $\endgroup$ Apr 13, 2012 at 9:07

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