Are locally presentable categories determined by their objects?

Question

Let $f: \mathcal{C} \rightarrow \mathcal{D}$ be a colimit-preserving functor between locally presentable categories. Assume that $f$ induces an equivalence between the groupoids underlying $\mathcal{C}$ and $\mathcal{D}$. Is $f$ necessarily an equivalence? What if $\mathcal{C}$ and $\mathcal{D}$ are assumed to be linear/dg/stable?

Answer 1

The answer in general is no.

Let $\mathcal C$ be the category of sets, let $\mathcal D$ be the category of pointed sets (with basepoint-preserving maps), and let $f: \mathcal C \to \mathcal D$ be the functor which adds a disjoint basepoint. Then $f$ is an equivalence on underlying groupoids, but not an equivalence of categories. Moreover, the forgetful functor is a right adjoint to $f$.

Of course, since any locally presentable 1-category is also presentable as an $\infty$-category, this answers the question in the $\infty$-categorical case, too. But for a more "intrinsically $\infty$-categorical" example, let $\mathcal C$ be the $\infty$-category of spaces, let $\mathcal D$ be the $\infty$-category of pointed spaces with disjoint basepoint (i.e. pointed spaces such that the connected component of the basepoint is contractible; morphisms are basepoint-preserving maps), and let $f: \mathcal C \to \mathcal D$ be the functor which adds a disjoint basepoint. Again, the forgetful functor is a right adjoint to $f$, and $f$ is an equivalence on object spaces. It's perhaps surprising that $\mathcal D$ is presentable, but it is: colimits are formed by taking the colimit in the category of pointed spaces and then collasping the connected component of the basepoint to be contractible; from this presentability is easy to verify.

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The answer in the stable case is yes.

Assume that $\mathcal C$ and $\mathcal D$ are stable (or even just additive with suspensions), and that $f$ is exact (or even just preserves finite direct sums and suspensions).

First, $f$ is faithful on the homotopy category. For if $f(\phi) = 0$, then $f\begin{pmatrix} 1 & \phi \\ 0 & 1 \end{pmatrix} = f\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $f$ is faithful on isomorphisms, we get $\begin{pmatrix} 1 & \phi \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, so that $\phi = 0$.

Now, $f$ is full on the homotopy category. For given a map $\phi$, because $f$ is full on isomorphisms, we have $f\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & \phi \\ 0 & 1 \end{pmatrix}$ for some $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, so that $f(b) = \phi$.

Thus $f$ is an equivalence on the homotopy category (since it is essentially surjective too). In this context, this implies that $f$ is an equivalence, as $\pi_n(Hom(X,Y)) = \pi_0(Hom(\Sigma^n X, Y))$.

This argument (without the bit about suspensions) also works for any additive ordinary category (i.e. any additive category with discrete hom-spaces).

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Here's a special case where the answer is yes.

Proposition: Let $f: L \to M$ be a sup-preserving map between complete lattices. Suppose that $f$ is a bijection. Then $f$ is an isomorphism.

Proof: We want to show that for $x,y \in L$, we have $x \leq y \Leftrightarrow f(x) \leq f(y)$. The forward implication follows from $f$ being sup-preserving (and hence order-preserving). For the reverse implication, suppose that $f(x) \leq f(y)$. We have $f(\sup(x,y)) = \sup(f(x),f(y)) = f(y)$. Because $f$ is a bijection, we have $\sup(x,y) = y$, i.e. $x \leq y$.

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Two observations for the general case: Let $f_!: \mathcal C \to \mathcal D$ be a left adjoint between locally presentable categories which induces an equivalence $\iota f_!: \iota \mathcal C \to \iota \mathcal D$ of spaces of objects, as in the question statement.

Claim: $f_!$ is conservative.

Proof: We may factor $f_!$ as a localization $l_!: \mathcal C \to L\mathcal C$ followed by a conservative left adjoint $g_!: L\mathcal C \to \mathcal D$. Because $l_!$ is a localization, its right adjoint $l^\ast$ is fully faithful. But this means that on object spaces, $\iota f_!$ factors through a retract $\iota L \mathcal C$ of $\iota \mathcal C$. Since $\iota f_!$ is an equivalence, $\iota l_!$ must be an equivalence, and $\iota l^\ast$ must also be an equivalence. Because $l^\ast$ is fully faithful, this implies that $l^\ast$ must be an equivalence, so that $l_!$ is also an equivalence. Thus $f_!$ may be identified with the conservative functor $g_!$.

Corollary: $f_!$ is faithful.

Proof: This is a general fact: a conservative left adjoint between cocomplete categories is faithful, for if $f_!(\phi) = f_!(\psi)$ for some $\phi,\psi: C^\to_\to C'$, then the map $f_! C' \to coeq(f_!\phi,f_!\psi)$ is an isomorphism. Since $f_!$ is conservative and preserves colimits, the map $C' \to coeq(\phi,\psi)$ is an isomorphism, i.e. $\phi = \psi$.

For the faithfulness statement, I'm not 100% sure how this translates into the $\infty$-categorical world.