Isomorphism preserves the operation of domain, so codomain inherits some properties of domain related to operation. So isomorphism is somewhat like "Equivalence". But, if the property of domain is not "related to operation", maybe that property is not preserved. So is there a property that is not preserved by isomorphism? Or Isomorphism is a "generalized equivalence"?

Thank you for your answer in advance, and I apologize for my terrible English.


closed as unclear what you're asking by Alex M., MickG, user147263, choco_addicted, Daniel W. Farlow May 14 '16 at 2:31

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  • $\begingroup$ The word "isomorphism" is used in context, such as an isomorphism of rings or an isomorphism of graphs. It means that all the properties in that context are preserved. There could be "properties" not preserved outside of the context for the word "isomorphism". For example, sometimes small groups have names, such as the "Klein four group". One might have a group isomorphism from a group with one name to a (structurally identical) group with a different name. $\endgroup$ – hardmath May 13 '16 at 22:07
  • $\begingroup$ Once the structures used to construct models are well defined in some meta language, the question what properties of the meta language are preserved under isomorphism is perfectly valid. Bourbaki, in the case of his structures, said that it is not hard to see which ones are preserved, but they have not characterized these properties. A few papers have been written about this question in the early times. Properties in the language itself are all preserved in all known logics, but it is not a tautology. Proofs must be written and different proofs have been written for different logics. $\endgroup$ – Dominic108 Nov 30 '18 at 23:28
  • $\begingroup$ I feel this question should not have been closed. It is ambiguous because we don't know which properties are considered, isomorphism of what kind of structures, etc. but, in this case, the ambiguity is what makes the question possible and it is an ambiguity that is very common. So, it deserves to be addressed as it is, with its inherent ambiguity, and can be answered. $\endgroup$ – Dominic108 Nov 30 '18 at 23:45

Yes an no.

No: The point of introducing the notion of an isomorphism is exactly that it should preserve all the relevant structure. If, say, $G$ and $G'$ are isomorphic groups, then every sentence you can write in the language of group theory will be true for $G$ if and only if the same sentence is true for $G'$ (i.e. you cannot distinguish between $G$ and $G'$ if your vocabulary includes only the group operation, variables, the identity element and basic logic). You might want to look up and introduction to category theory for more details.

Yes: If you go outside the language relevant to the isomorphism you have in mind, then surely you can find lots of properties which will not be preserved. The groups $G = \mathbb{Z}/4\mathbb{Z}$ and $G' = (\{1,i,-1,-i\}, \cdot)$ are isomorphic, but one of then contains $i$ while the other one does not (it does contain the equivalence class $1 + 4 \mathbb{Z}$, but that's not quite the same thing). Of course, this is a trivial example, but in general isomorphism of groups will preserve the properties that apply to groups, but not any other type of properties.

Edit to add: It is worth also pointing out that there are various species of isomorphisms: group isomorphisms, isomorphisms of topological spaces (aka homeomorphisms), isomorphisms of topological groups, and so on.

Consider a triangle and a disc: they are isomorphic as topological spaces, but not as metric spaces (i.e. you can continuously deform a triangle into a circle, but you cannot do so while preserving distances between pairs of points).

Another example which I like is to take two different representations of the additive group $\mathbb{R}$: one is just the usual one, and the other one is $G = \{(t \bmod 1,\sqrt{2}t \bmod 1) \ : \ t \in \mathbb{R} \}$, equipped with addition modulo $1$. Then $(\mathbb{R},+)$ and $(G,+)$ are isomoprhic as groups. However, as topological spaces they look very different!

  • 2
    $\begingroup$ Another example people can potentially trip over is that the map $\mathbb{Z}\mapsto 2\mathbb{Z}$ by $x \mapsto 2x$ is an isomorphism of the additive groups, but it's not an isomorphism of pseudo-rings (indeed, there isn't one, since $\mathbb{Z}$ has a multiplicative identity and $2\mathbb{Z}$ doesn't). Of course, if you equip the set of even numbers with a different multiplication operation from the one that $2\mathbb{Z}$ has, then you could come up with a ring isomorphism. So when someone talks about morphisms and it's not obvious which structure they mean, they need to say. $\endgroup$ – Steve Jessop May 13 '16 at 18:22
  • $\begingroup$ How can you tell $i \in G'$ and $1 + 4 \mathbb Z \in G$ apart? $\endgroup$ – Earthliŋ May 13 '16 at 20:18
  • $\begingroup$ @Earthliŋ: Well, these are just different things: one is a number in the complex plane, the other is a set of integers. It's like apples and oranges: an apple is not an orange and orange is not an apple - even though you count them in the same way. $\endgroup$ – Jakub Konieczny May 13 '16 at 21:29
  • $\begingroup$ You mean you look at the underlying sets, I see. $\endgroup$ – Earthliŋ May 13 '16 at 21:31

This question is tautological, insofar that an isomorphism of a structure, by definition, preserves that structure.

That said, there are basically two potential issues here:

The first is that one can have an isomorphism between two structures on different underlying sets. A simple exactly is the permutation group $S_{\{A, B, C\}}$ on the letters $A, B, C$ and the group of symmetries $T$ of an equilateral triangle. One can define a group isomorphism $\phi : S_{\{A, B, C\}} \to T$ by labeling the corners of the triangle by $A, B, C$, and then declaring the image of a permutation to be the unique symmetry of the triangle that permutes the vertices according to the labels. Of course, the elements of $S_{\{A, B, C\}}$ are not those of $T$, so $\phi$ does not preserve the 'property' of membership in $S_{\{A, B, C\}}$.

Second, it is certainly possible to have an object with some structure and a map that preserves some of the structure but not all of it. A simple example: Define the partially ordered set $S := (\{0, 1\}, \prec )$, where declare $0 \prec 1$. Then, the map $\phi : S \to S$ defined by $\phi(0) = 1, \phi(1) = 0$ is an isomorphism of sets (this just means that $\phi$ is a set bijection), but not an isomorphism of partially ordered sets, as $0 \prec 1$ but $\phi(0) \not\prec \phi(1)$. (Of course, since $\phi$ is a map $S \to S$, it does preserve the 'property' of membership of $S$.)

The latter distinction can sometimes be subtle: For example, there is an isomorphism between $\Bbb R$ and $\Bbb R^2$ as vector spaces over $\Bbb Q$, but no such map is an isomorphism of real vector spaces, as $\Bbb R$ and $\Bbb R^2$ have different dimensions as real vector spaces.


If I understand it correctly, your question contains its own answer: consider the groups $(\Bbb R, +)$ and $\Big( (0, \infty), \cdot \Big)$, and the isomorphism $\exp : \Bbb R \to (0, \infty), \ \exp(x) = \textrm e ^x$. Choose now a property that has nothing to do with the algebraic operation, for instance choose some topological property: endow $\Bbb R$ with the trivial topology $(\emptyset, \Bbb R)$ and $(0, \infty)$ with the discrete topology. Then $\Bbb R$ with this topology will be compact and non-Hausdorff, while $(0, \infty)$ with the discrete topology will be non-compact and Hausdorff. Despite these tremendous topological differences, the two groups are algebraically isomorphic. So yes, there do exist properties that are not preserved by isomorphisms.


Names, basically. Otherwise an isomorphism indicates that the structures are identical (/compatible). If two (sets/groups/rings/...) are isomorphic, you can basically just think of them as being the same structure, with differently-named elements.


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