Existence of a certain functor $F:\mathrm{Grpd}\rightarrow\mathrm{Grp}$ 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

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.
