# Definition of “closure under isomorphism”

What does " Closed Under Isomorphism " means ? why do we need it ? And how can we use it in " Mathematical Structures" thanks

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I have never heard that phrase used before. – Dustan Levenstein Mar 18 '12 at 17:08
Now you heard about it :) – M.M Mar 18 '12 at 17:15
Why is this tagged (data-analysis)? What do you mean by mathematical structure? Can you provide definitions? – user2468 Mar 18 '12 at 17:24
@Mohammed: It would help if you could provide an excerpt or link to the place where this terminology was used. – Pete L. Clark Mar 18 '12 at 18:00
www.cs.tau.ac.il/~nachumd/ASM/Nondeterminism.pptx slid 3 – M.M Mar 18 '12 at 18:11

Since nobody answered this question I will give it a try. I encountered closure under isomorphism in the definition of a Markov property in group theory (doesn't correspond to the Markov property I find on Wikipedia). One of the requirements for a property of groups $\mathcal{B}$ to be a Markov property is that it is closed under isomorphism $$\mathcal{B}\text{ is true for a group }G\Leftrightarrow\mathcal{B}\text{ is true for every group isomorphic to }G$$ Most (arguably all) properties one considers in group theory (finite, abelian, ...) are closed under group isomorphism.

So I would generally define a property $\mathcal{B}$ of objects in a category $\mathcal{C}$ to be closed under isomorphism iff $$\mathcal{B}\text{ is true for }A\in\mathcal{C}\Leftrightarrow\mathcal{B}\text{ is true for every object in }\mathcal{C}\text{ isomorphic to }A$$

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Isn't this more commonly called "invariant under isomorphism" or "preserved under isomorphism"? – jwodder Mar 18 '12 at 20:50
@jwodder: True, I would probably also use "preserved under isomorphism". Andrew Glass, who taught a course on decision problems in group theory last year, used "closed under isomorphism". – Michalis Mar 18 '12 at 21:09

My first reaction to the phrase is that it would be used in a context such as

The category $\mathcal{C}$ is a subcategory of $\mathcal{D}$. We say that $\mathcal{C}$ is closed under isomorphism iff:

If $f : X \to Y$ is an arrow of $\mathcal{C}$ and $g : Y \to Z$ and $h : W \to X$ are both isomorphisms in $\mathcal{D}$, then $gf$ and $fh$ are both arrows of $\mathcal{C}$.

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