(non-)Existence of an endofunctor of $F$ on $g/{\bf B}G^\circlearrowleft$ given the definition of $F$ on objects.

Question

In few words: given an element of a group $g\in G$ given an I'd like to define and endofunctor $$F:g/{\bf B}G^\circlearrowleft\to g/{\bf B}G^\circlearrowleft$$ I have defined $F$ on objects and I'd like to extend it to arrows obtaining functoriality (here $g^h=hgh^{-1}$). $$F_{Ob}:(g\overset{\alpha}{\to}g^\alpha)\mapsto (g\overset{g^\alpha}{\to}g^{g^\alpha})$$ I tried to find a possible definition for $F_{Arr}$. Clearly setting $F_{Arr}(\phi):=g^\phi$ doesn't work not even at a domain/codomain level, let alone full functoriality. I tried to enforce the functoriality conditions on $F_{Arr}(\phi):=g^\phi$ but I'm lost in a ocean of parentheses so I gave up atm. I strongly suspect that no possible definition of $F_{Arr}$ makes $F=(F_{Ob},F_{Arr})$ an endofunctor.

Question 1: It is possible to extend $F_{Ob}$ to an endofunctor of $g\setminus{\bf B}G^\circlearrowleft$? If yes show a possible definition for $F_{Arr}$.

Question 2: If funtoriality is not possible can we tweak $F_{Ob}$ so to obtain an endofunctor of $g/{\bf B}G^\circlearrowleft$ (or some variant of it) s.t. for a fixed $g$ it maps in some way $$h\to g^h=hgh^{-1}$$

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Notation: Let $G$ be a group then ${\bf B}G$ is a category with a single object (delooping) $\bullet$ and with ${\bf B}G(\bullet,\bullet):=G$.

Define ${\bf B}G^\circlearrowleft:=[{\bf B}\mathbb N,{\bf B}G]$ i,e. the category of endomorphisms of ${\bf B}G$ and commutative squares between them. In other words its objects are morphisms $g^-:\mathbb Z\to G$ mapping $n\mapsto g^n$ that we know are in bijection with elements of $G$ and as arrows $\alpha:g\Rightarrow h$ natural transformations, i.e. elements $\alpha \in G$ s.t. $\alpha g=h\alpha$ (modulo some technicality).

Fixed a $g\in G$ we define $g/{\bf B}G$ to be the coslice category (undercategory):

• objects of the category $$g\overset{\alpha}{\Rightarrow}h$$ are pairs $(\alpha,h)=(\alpha,g^\alpha)$, i.e. $\alpha g=h\alpha $;
• Arrows $$\phi:(\alpha,h)\to(\beta,f)$$ are morphisms $\phi: h\Rightarrow f$, i.e. $\phi h=f\phi$, that in addition satisfies $\phi \alpha=\beta$ so we get

$$\phi:(\alpha,h)\to(\phi\alpha,h^\phi)$$

And this is consistent with what we know of the objects. In fact we get that the domain of every morphism is completely determined by its domain and by $\phi$ itself.

$$\phi:(\alpha,g^\alpha)\to (\phi\alpha, (g^\alpha)^\phi)$$

This last detail must be crucial for my question but my brain is a bit foggy on this.

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More context/attempt at setting up the problem

Now fix $g\in G$ and define $\Sigma:G\to G$ to be $\Sigma(h)=g^h$. We have $\Sigma(g)=g$ and $\Sigma(\phi h)=\Sigma(h)^\phi$ (idk if useful). Note that every object of $g/{\bf B}G^\circlearrowleft$ is of the form $(\alpha,\Sigma(\alpha))$

Define $F_{Ob}$ as the component on object of the desired functor.

$$F_{Ob}(\alpha,\Sigma(\alpha)):=(\Sigma(\alpha),\Sigma^2(\alpha))$$ It is still an object of the category.

How to define $F_{Arr}(\phi)$? Let $\phi:(\alpha, \Sigma(\alpha))\to(\beta,\Sigma(\beta))$ we look for $$F_{Arr}(\phi):(\Sigma(\alpha), \Sigma(\Sigma(\alpha)))\to(\Sigma(\beta),\Sigma(\Sigma(\beta)))$$ $$F_{Arr}(\phi):(\Sigma(\alpha), \Sigma(\Sigma(\alpha)))\to(\Sigma(\phi\alpha),\Sigma(\Sigma(\phi\alpha)))$$

So we should have $F_{Arr}(\phi)\Sigma(\alpha)=\Sigma(\beta)=\Sigma(\phi\alpha)$. I believe that this condition completely determine the component on arrows as

$$F_{Arr}(\phi)=\Sigma(\phi\alpha)\Sigma(\alpha)^{-1}=g^{\phi\alpha}(g^{-1})^{\alpha}$$

But do we have $\Sigma^2(\alpha)^{F_{Arr}(\phi)}=\Sigma^2(\phi\alpha)$? Does $F_{Arr}(\phi\psi)=F_{Arr}(\phi)F_{Arr}(\psi)$? I believe that $F_{Arr}$ has to be an endomorphism of $G$.

At this point I'm lost. Is it functorial? Maybe it is obvious but I worked non stop on this all the night. I just see a mes of parentheses. I don't believe it but I don't have a proof. If it isn't can we fix the definition to obtain functoriality?

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Update

I'm convinced that I found a proof. I made my answer community wiki. I'm not going to delete the question because maybe some better proof can be proposed or a mistake in my proof can be found. I also improve the tags to reflect the group theoretic origin of this construction.

Answer 1

The answer was straightforward. I was just few computations away but lacking the right organization of the notation.

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Define $F:g/{\bf B}G^\circlearrowleft\to g/{\bf B}G^\circlearrowleft$.

• $F_{\rm Ob}:(\alpha, h)\mapsto(\Sigma(\alpha),\Sigma(h))$;
• $F_{\rm Arr}:(\alpha,h)\overset{\phi}{\longrightarrow}(\beta,f)\quad\mapsto\quad(\Sigma(\alpha),\Sigma(h))\overset{\Sigma(\phi\alpha)\Sigma(\alpha)^{-1}}{\longrightarrow}(\Sigma(\beta),\Sigma(f))$;

Proposition. $F$ is a functor.

Proof:

• (identity) $F_{\rm Arr}({\rm id}_{(\alpha,h)})=F_{\rm Arr}((\alpha,h)\overset{1_G}{\longrightarrow}(\alpha,h))=(\Sigma(\alpha),\Sigma(h))\overset{1_G}{\longrightarrow}(\Sigma(\alpha),\Sigma(h)) \\{\rm id}_{F_{\rm Ob}((\alpha,h))}={\rm id}_{F_{\rm Ob}((\Sigma(\alpha),\Sigma(h)))}=(\Sigma(\alpha),\Sigma(h))\overset{\Sigma(\alpha)\Sigma(\alpha)^{-1}}{\longrightarrow}(\Sigma(\alpha),\Sigma(h))$:

• (domain) $F_{\rm Ob}({\rm dom}((\alpha,h)\overset{\phi}{\longrightarrow}(\beta,f)))=F_{\rm Ob}((\beta,f))=(\Sigma(\beta),\Sigma(f))\\ {\rm dom}(F_{\rm Arr}((\alpha,h)\overset{\phi}{\longrightarrow}(\beta,f)))={\rm dom}((\Sigma(\alpha),\Sigma(h))\overset{\phi}{\longrightarrow}(\Sigma(\beta),\Sigma(f)))=(\Sigma(\beta),\Sigma(f))$

• (codomain) Same goes for the codomain

• (composition) Take two composable morphisms in $g/{\bf B}G^\circlearrowleft$ $$(\alpha,h_0)\overset{\phi}{\rightarrow}(\beta,h_1)\overset{\psi}{\rightarrow}(\gamma,h_2)$$ rewrite them using the definition of arrows in coslice categories $$(\alpha,h_0)\overset{\phi}{\rightarrow}(\phi\alpha,h_1)\overset{\psi}{\rightarrow}(\psi\phi\alpha,h_2)$$ rewrite again using the definition of arrows in ${\bf B}G^\circlearrowleft$ $$(\alpha,\Sigma(\alpha))\overset{\phi}{\rightarrow}(\phi\alpha,\Sigma(\phi\alpha))\overset{\psi}{\rightarrow}(\psi\phi\alpha,\Sigma(\psi\phi\alpha))$$ Apply $F$ $$(\Sigma(\alpha),\Sigma^2(\alpha))\overset{\Sigma(\phi\alpha)\Sigma(\alpha)^{-1}}{\longrightarrow}(\Sigma(\phi\alpha),\Sigma^2(\phi\alpha))\overset{\Sigma(\psi\phi\alpha)\Sigma(\phi\alpha)^{-1}}{\longrightarrow}(\Sigma(\psi\phi\alpha),\Sigma^2(\psi\phi\alpha))$$ and this means that $$F_{\rm Arr}(\psi)F_{\rm Arr}(\phi)=(\Sigma(\psi(\phi\alpha))\Sigma(\phi\alpha)^{-1})(\Sigma(\phi\alpha)\Sigma(\alpha)^{-1}) \\ F_{\rm Arr}(\psi\phi)=\Sigma((\psi\phi)\alpha)\Sigma(\alpha)^{-1}$$ $\square$