Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G, G'$ be two groups and $X=\{x,y\}$ be a set of two elements. Consider a groupoid $\mathcal{G}$ with objects from $X$ such that Hom$(x,x)=G$ and Hom$(y,y)=G'$.

Suppose Hom$(x,y) \neq \emptyset$, i.e. there is a morphism between $x$ and $y$. I think this is possible if and only if this morphism is an isomorphism between $G$ an $G'$ (in fact, every morphism in a groupoid must be invertible, moreover it must respect multiplication, so it is an isomorphism of these groups). Also, I know that any two such morphisms are conjugate.

I need to prove: "Given two groupoids $\mathcal{G}_1$ and $\mathcal{G}_2$ satisfying the conditions above (i.e. both have $X$ as set of objects and $G, G'$ as hom-sets Hom$(x,x)$ and Hom$(y,y)$), if there is a morphism from $x$ to $y$ in both groupoids, then $\mathcal{G}_1$ and $\mathcal{G}_2$ are isomorphic".

[Two groupoids are isomorphic if they are isomorphic as category, i.e. there is an "invertible" functor between the two].

It seems quite an obvious statement, but I cannot prove it. I start with a functor between the groupoids, say $F: \mathcal{G}_1 \to \mathcal{G}_2$. Then, what? Should I concretely construct such a functor and show it is an isomorphism? Or is there an "abstract nonsensical way" to do it?

share|improve this question
    
If you don't specify morphisms between the objects in your groupoid when you construct it, there are none. –  Tobias Kildetoft Mar 12 '13 at 15:54
    
In the groupoid you constructed in your first paragraph there are no morphisms from $x$ to $y$, so it is not clear what you mean by «suppose now that $\hom(x,y)\neq\emptyset$...» in the second paragraph! –  Mariano Suárez-Alvarez Mar 12 '13 at 15:55
    
Ok, I will edit the question to make clear what I mean. Sorry! –  user39280 Mar 12 '13 at 15:56
    
I slightly changed the question, I hope now it is clearer what the problem is. –  user39280 Mar 12 '13 at 15:59
1  
You were still using the word "construct" which is what they objected to. I've edited your question to say what I think you meant. If I've gotten it wrong feel free to rollback. –  Jim Mar 12 '13 at 16:06
show 1 more comment

2 Answers 2

up vote 1 down vote accepted

You should just construct the functor. The objects map to themselves as do the endomorphisms. All you really have to decide is how the map $F\colon\hom_{\mathcal G_1}(x, y) \to \hom_{\mathcal G_2}(x, y)$ is going to work.

Pick $\phi_1 \in \hom_{\mathcal G_1}(x, y)$ and $\phi_2 \in \hom_{\mathcal G_2}(x, y)$. What you need to do is show that every $f \in \hom_{\mathcal G_1}(x, y)$ can be written in the form $\phi_1f'$ where $f' \in G$, and similarly for $\mathcal G_2$. Then define $F(\phi_1f') = \phi_2f'$.

You'll have to show that $F$ respects composition and is a bijection on that homset, but that shouldn't be hard.

share|improve this answer
    
Thank you. Very clear! I knew the answer was not particularly difficult but I could not come up with anything. –  user39280 Mar 12 '13 at 16:16
add comment

Let $\mathcal{G}_i$ has the objects $x_i,y_i$ with the automorphism groups $G_i,H_i$. Fix isomorphisms $\phi:G_1\to H_1$, $\psi:G_2\to H_2$, $\alpha:G_1\to G_2$. Now build a functor $F$: for $f\in G_1$ define $Ff=\alpha f\alpha^{-1}$ and for $g\in G_2$ define $Fg=\psi\alpha\phi^{-1} g \phi\alpha^{-1}\phi^{-1}$, and so on.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.