TITLE: Is there a property that is not preserved by isomorphism?
QUESTION [23 upvotes]: 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.

REPLY [7 votes]: 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.