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Let $\mathrm{Grpd}$ denote the category of all groupoids. Let $\mathrm{Grp}$ denote the category of all groups. Are there functors $F\colon\mathrm{Grpd}\rightarrow \mathrm{Grp}, G\colon\mathrm{Grp}\rightarrow \mathrm{Grpd}$ such that $GF=1_{\mathrm{Grpd}}$.

Dear all, I know the question is not easy (at least for me). I don't expect you to solve a problem that is possibly unintresting for you and waste your time on it. I was just asking to see if anyone had seen something similar so that s/he would give me a reference to it

Thank you

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    $\begingroup$ @Kevin Carlson I tried to find such a functor but I couldnt. I also don't think that there is anything worth stating in my trials to find such a functor $\endgroup$ – Amr May 1 '13 at 23:19
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    $\begingroup$ I have no reason to expect such a pair of functors to exist. You should at least indicate what you think the answer is and why! $\endgroup$ – Zhen Lin May 1 '13 at 23:22
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    $\begingroup$ That's fine-you could also say something about why you want such functors and the amount of category theory you have. In particular, the requirement that $GF=1$ instead of $GF\cong 1$ is unnatural. $\endgroup$ – Kevin Carlson May 1 '13 at 23:23
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    $\begingroup$ @KevinCarlson Even allowing for $G F \cong \mathrm{id}$ instead of $G F = 1$ doesn't really change my expectation that no such pair of functors exist. For instance, where are the discrete groupoids going to go, and how are we going to get them back? $\endgroup$ – Zhen Lin May 1 '13 at 23:25
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    $\begingroup$ Dear all, I know the question is not easy (at least for me). I don't expect you to solve the problem and waste your time on it. I was just asking to see if anyone had seen something similar S/he would give me a refrence to it $\endgroup$ – Amr May 1 '13 at 23:33
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Such a pair of functors do not exist.

Reason 1 (if you accept the empty groupoid)

In the category of groups every pair of objects have a morphism between them. While in the the category of groupoids there is no morphism from the terminal object to the initial object. It follows that the initial object can't be in the image of $G$.

Reason 2 (if you don't accept the empty groupoid)

Let $A$ and $B$ be the discrete groupoids on the sets $\{0\}$ and $\{0,1\}$ respectively, and let $f,g:A\to B$ be functors defined by $f(0)=0$ and $g(0)=1$. Now suppose such a pair of functors exists and let $z: F(A)\to F(A)$, $z':F(A)\to F(B)$ be the group homorphisms sending everything to the identity element. Since $A$ is the terminal object in the category of groupoids it follows that $G(z) = 1_A$. We have $f = f 1_A= GF(f) G(z)= G(F(f)z)=G(z')$ and similary $g= G(z')$. This leads to $f=g$ which is a contradicition.

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