# On why equivalent categories $E : I \simeq J$ have equivalent categories of cones $\int \mathbf{Cone}(\_,F) \simeq \int\mathbf{Cone}(\_,FE)$

I'm currently working on the following exercise from Emily Riehl's Category Theory in Context,

Exercise 3.1.xii. Suppose $$E:I \stackrel{\simeq}{\to} J$$ defines an equivalence between small categories and consider a diagram $$F : J \to C$$. Show that the category of $$J$$-shaped cones over $$F$$ is equivalent to the category of $$I$$-shaped cones over $$FE$$, and use this equivalence to describe the relationship between limits of $$F$$ and limits of $$FE$$.

Here $$J$$ is assumed to be small and $$C$$ locally small, although I am not sure this assumptions are needed. I have found a similar idea to mine here, which is to define the following functor

\begin{align} \Gamma : \int &\mathbf{Cone}(\_,F) \longrightarrow \int\mathbf{Cone}(\_,FE) \\ & (c,(\lambda_j)_{j\in J}) \longmapsto (c,(\lambda_{Ei})_{i\in I}) \\ (f:c \to d &\text{ s.t. }\lambda_jf = \mu_j) \longmapsto (f:c \to d \text{ s.t. } \lambda_{Ei}f = \mu_{Ei}) \end{align}

that takes $$\lambda : c \Rightarrow F$$ to the natural transformation that has its components for each object $$Ei$$, and takes a morphism that commutes with the cones to itself (since it will still commute with the selected legs). Now, it is asserted in the linked post that $$\Gamma$$ should be an isomorphism of categories (in particular an equivalence), but the details aren't specified and I haven't managed to finish the job. So far, here is what I have come up with:

$$\Gamma$$ is (essentially) surjective: take a cone $$\nu : c \Rightarrow FE$$. For each $$j \in J$$, since $$E$$ is essentially surjective take $$\varphi_j : Ei_j \xrightarrow{\sim} j$$ be an isomorphism. Now if we define $$\mu_j$$ to be the composite $$c \xrightarrow{\nu_{i_j}} FEi_j \xrightarrow{F\varphi_j} Fj$$ then $$\mu$$ is a cone over $$c$$: if $$f: j \to j'$$ is an arrow on $$J$$, then $$\varphi_{j'}^{-1}f\varphi_j: Ei_j \to Ei_{j'}$$ must come from a (unique) arrow $$s : i_j \to i_{j'}$$ such that $$Es = \varphi_{j'}^{-1}f\varphi_j$$. Consequently, \begin{align} Ff\mu_j &= FfF\varphi_j\nu_{i_j} = F(f\varphi_j)\nu_{i_j} = F\varphi_{i_{j'}}(FEs)\nu_{i_j} \\ & =F\varphi_{i_{j'}}(FEs)\nu_{i_j} = F\varphi_{i_{j'}}\nu_{i_{j'}} = \mu_{j'} \end{align} and so in effect $$\mu$$ is a cone over $$c$$. Since we can in particular take each $$\varphi_{Ei}$$ to be $$1_{Ei}$$ for each $$j$$ such that $$j = Ei$$, we get that $$\Gamma(c,\mu) = (c,\nu)$$.

It suffices to see now that $$\Gamma$$ is fully faithful (to prove the equivalence, at least). It is clear to me that if $$f,g: c \to d$$ are distinct morphisms that commute with the respective cones over $$c$$ and $$d$$, by construction we get that $$\Gamma f = f \neq g = \Gamma g$$.

However, if we have $$f: c \to d$$ such that it commutes with the legs of the cones that correspond to the image of $$E$$, why should it be that $$f$$ commutes with all legs of the original cones?

Let $$f:c\to d$$ be such that it commutes with the cones in the image of $$E$$; and let $$j\in J$$.

Let $$g:j\to Ei$$ be an isomorphism. You have the following commutative diagrams (sorry for the formatting, I don't know how to do triangles on here)

$$\require{AMScd} \begin{CD} c @>{id_c}>> c\\ @V{\lambda_j}VV @VV{\lambda_{Ei}}V\\ Fj @>>{Fg}> FE_i \end{CD}$$ because $$(c,\lambda)$$ is a cone

$$\require{AMScd} \begin{CD} d @>{id_d}>> d\\ @V{\mu_{Ei}}VV @VV{\mu_j}V\\ FEi @>>{Fg^{-1}}> Fj \end{CD}$$ because $$(d,\mu)$$ is a cone

$$\require{AMScd} \begin{CD} c @>{f}>> d\\ @V{\lambda_{Ei}}VV @VV{\mu_{Ei}}V\\ FEi @>>{id_{FEi}}> FE_i \end{CD}$$ by hypothesis

Then you can put these three diagrams side by side in the order (1-3-2) to get

$$\require{AMScd} \begin{CD} c @>{f}>> d\\ @V{\lambda_j}VV @VV{\mu_j}V\\ Fj @>>{id_{Fj}}> Fj \end{CD}$$ because $$id_d\circ f\circ id_c = f, Fg^{-1}\circ id_{FE_i}\circ Fg = id_{Fj}$$

and this is exactly what you want. So this comes down to the facts that $$(c,\lambda)$$, $$(d,\mu)$$ are cones over all $$J$$, that $$F$$ is a functor and that $$E$$ is essentially surjective.

• Can I ask what field of mathematics is this? It looks interesting Dec 25, 2018 at 15:29
• As usual, awesome answer! Thanks a lot for taking the time to answer. Dec 25, 2018 at 15:31
• @KKZiomek : this is category theory; the OP mentions an interesting textbook about this topic ! Guido A. : you're welcome Dec 25, 2018 at 15:34