Equivalence of category of cones If $ E \colon I \rightarrow J $ is an equivalence of categories and $ D \colon J \rightarrow C $ is a diagram of shape $ J $ in $ C $, is the category of cones over $ D $ equivalent to the category of cones over $ D \circ E $?
 A: An elementary solution can be made to work, but let me suggest a slightly more sophisticated route. The first observation is that the category of cones over a diagram $D$ is the category of functors $\text{Func}(\text{cone}(J),C)_D$ from the cone category over $J$ to $C$ which restrict to $D$ along the inclusion $J\to \text{cone}(J)$. Here $\text{cone}(J)$ is just $J$ with a new initial object adjoined. It's straightforward to show that $E$ induces an equivalence between $\text{cone}(I)$ and $\text{cone}(J)$ (when extended to map the new initial objects to each other.) Thus the categories $\text{Func}(\text{cone}(I),C)$ and $\text{Func}(\text{cone}(J),C)$ of all cones over $I$-and $J$-shaped diagrams are equivalent, being functor categories out of equivalent categories. This equivalence sends a cone over $D$ to the corresponding cone over $D\circ E$, so restricts to an equivalence of $\text{Func}(\text{cone}(J),C)_D$ with $\text{Func}(\text{cone}(I),C)_{D\circ E}$
This is still a bit of a mouthful, so a yet more sophisticated and shorter proof would interpret $\text{Func}(\text{cone}(J),C)_D$ using a touch of 2-category theory. Briefly, $\text{cone}(J)$ is a pushout in the 2-category of categories and similarly $\text{Func}(\text{cone}(J),C)_D$ is an equalizer, so the 2-categorical analogue of the fact that limits and colimits of isomorphic diagrams are isomorphic implies that the two cone categories are equivalent.
