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?


1 Answer 1


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.

  • $\begingroup$ Can I ask what field of mathematics is this? It looks interesting $\endgroup$
    – KKZiomek
    Dec 25, 2018 at 15:29
  • $\begingroup$ As usual, awesome answer! Thanks a lot for taking the time to answer. $\endgroup$
    – qualcuno
    Dec 25, 2018 at 15:31
  • 2
    $\begingroup$ @KKZiomek : this is category theory; the OP mentions an interesting textbook about this topic ! Guido A. : you're welcome $\endgroup$ Dec 25, 2018 at 15:34

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