Equivalence of certain comma categories

Let $\mathcal{C}$ be some category, $A \in \mathcal{C}$, $\mathbf{Grpd}_A$ be a groupoid consisting of all objects isomorphic to $A$ and all isomorphisms between them, and $i: \mathbf{Grpd}_A \to \mathcal{C}$ be the embedding. Are categories $\mathcal{C} \downarrow A$ and $\mathcal{C} \downarrow i$ necessarily equivalent?

I tried to construct the equivalence for a special case of bundles but there was an obstruction (one triangle that failed to be commutative), and I don't know how to approach the proof of inequivalence in general. Should I seek to prove inequivalence directly for some toy category (which I'd prefer not to do, see below) or is there an actual equivalence?

My motivation lies with bundles (so I'm mostly concerned with the special case $\mathcal{C} = \mathbf{Top}$): it seems 'evil' to fix one space as the base because we don't generally distinguish between isomorphic spaces, so a natural question arises: does it matter if we fix just one or the whole isomorphism class?

A hint would be nice, but if the question is actually too answer with only little knowledge of category theory (up to limits), a full answer is welcome too.

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Is the issue that you can't pick a coherent system of isomorphisms from objects in the groupoid to $A$? –  Dylan Moreland Aug 26 '11 at 23:39
@Dylan: arrange $p': E' \to B'$, $p'': E'' \to B''$ together with $u: E' \to E''$ and $f: B' \to B''$ as a convex trapezoid. Then add $\varphi': B' \to B$ and $\varphi'': B'' \to B$ so that arrows form a large triangle. That's how I tried to construct one functor of the equivalence (the other being, of course, the embedding). However, there is no guarantee that the small triangle in the diagram commutes, and this later came back to bite me when I was constructing a natural transformation to $\mathrm{id}_{\mathcal{C} \downarrow i}$. –  Alexei Averchenko Aug 26 '11 at 23:40
@Dylan: yes, and I'm not sure if picking such a system even fits the spirit of what I'm pondering :) –  Alexei Averchenko Aug 26 '11 at 23:41

These two categories are not equivalent in general, because $\mathcal{C}\downarrow A$ always has a terminal object $id: A \to A$ whereas $\mathcal{C}\downarrow i$ may fail to have one.

Assume that your original category $\mathcal{C}$ has an initial object $0$ (this will in particular apply to the case $\mathcal{C}=Top$), and consider any object $p: E \to B$ in $\mathcal{C}\downarrow i$.

Suppose that $A$ has some automorphism different from the identity. Then the same holds for $B$. This nonidentity automorphism $u: B \to B$ together with $id: B \to B$ will provide two different maps from $(0 \to B)$ to $p$. Consequently, $p$ cannot be a terminal object in $\mathcal{C}\downarrow i$.

In fact, if $A$ does not have any automorphisms beside the identity then $Grpd_A$ has exactly one isomorphism between any two of its objects and your attempt will then provide an equivalence.

As to your general question, I do not think that the choice of a particular object $B$ out of $Grpd_A$ runs against the categorical spirit. The important thing is that the result, the category $\mathcal{C}\downarrow B$, does not depend on this choice: any map $f: B \to B'$ induces a functor $F: \mathcal{C}\downarrow B \to \mathcal{C}\downarrow B'$ and if $f$ is an isomorphism then $F$ will be an equivalence. Therefore one may as well use $\mathcal{C}\downarrow A$ right from the beginning.

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