Call a category $C$ *rigid* if every equivalence $C \to C$ is isomorphic to the identity. I don't know if this is standard terminology. Many of the usual algebraic categories are rigid, for example sets, commutative monoids, groups, abelian groups, commutative rings, but also the category of topological spaces. The category of monoids (or rings) is not rigid because $M \mapsto M^{\mathrm{op}}$ is an equivalence which is not isomorphic to the identity. See [M] for a survey and the general strategy for proving rigidity. The case of commutative rings was discussed recently on MO here. The philosophy is that a category is rigid if every object can be defined in a categorical way, which is a quite interesting property.

**Question**: Is the category of schemes rigid?

Here is what I've done so far: The initial scheme is $\emptyset$ and the terminal scheme is $\text{Spec}(\mathbb{Z})$. Spectra of fields are characterized by the property that they are non-initial and and every morphism from a non-initial object to them is an epimorphism, see Kevin's answer here. The underlying set $|X|$ of a scheme is the set of equivalence classes of morphisms $Y \to X$, where $Y$ is the spectrum of a field. So this recovers $|X|$ from $X$ in a categorical manner. If $x \in |X|$, then $\text{Spec}(\kappa(x))$ is the terminal spectrum of a field which maps to $X$ and has (set) image $x$.

However, I'm not able to recover the topology from $X$. I don't know how to characterize open or closed immersions. They are exactly the étale resp. proper monomorphisms, see this MO question, but it seems to be hard to characterize étale and proper categorically. After all, if are able to characterize affine schemes, then we will be done, since the category of affine schemes is rigid and every scheme is the canonical colimit of the affine schemes mapping into it.

In order to characterize affine schemes, it is enough to characterize the ring object $\mathbb{A}^1_\mathbb{Z}$ in the category of schemes, since we can then define the ring of global sections of a scheme categorically and then say that affine schemes $Y$ are characterized by the property that for all schemes $X$ the map $Hom(X,Y) \to Hom(\mathcal{O}(Y),\mathcal{O}(X))$ is bijective.

Other approaches: 1. First show that the category of fields is rigid. I've already shown that the notions of prime field, $\mathbb{F}_p$, $\mathbb{Q}$, finite, characteristic, normal, separable, algebraic, galois, transcendent, transcendence degree are categorical, but this is not enough to distinguish, for example, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. If $F$ is a self-equivalence of the category of fields, then $F$ maps $K(X)$ to $F(K)(X)$, so taking automorphisms there is a natural isomorphism $\text{PGL}(2,K) \cong \text{PGL}(2,F(K))$, but I wonder if this already implies that $K \cong F(K)$ naturally. 2. Characterize local schemes as a special full reflective subcategory containing the spectra of fields. 3. Try to categorify cohomology theory and use Serre's criterion for affineness.

EDIT (May '11): I've restarted this project in the last days. If $k$ is a field with only trivial endomorphisms, then I can show that every self-equivalence of $\text{Sch}/k$ preserves $\text{Spec}(k[\epsilon]/\epsilon^2)$, but also $\text{Spec}(k[[t]])$. But I still have no idea how to approach $\text{Spec}(k[t])$ categorically. Even basic notions such as "closed point" or "quasicompact" remain unclear.

EDIT (Feb '12): Let's work with $\mathrm{Sch}/k$ for some algebraically closed field $k$. Then $F$ maps $\mathbb{A}^1_k$ to a ring object in $\mathrm{Sch}/k$. If we already knew that it is of finite type over $k$ and irreducible, then a Theorem by Greenberg (Cor. 4.4 in *Algebraic Rings*, Trans. AMS, Vol. 111, No. 3, pp. 472 - 481) will imply that the underlying scheme is just $\mathbb{A}^n_k$ for some $n$. Now using my question about factorization we should be able to conclude $n=1$. Of course, many details are missing here; for example it is not clear at all why $F$ should preserve schemes of finite type.

Any ideas concerning the categorical characterization of other properties / objects are appreciated. Feel free to add every piece as a single answer even if it does not answer the whole question.

[M] E. Makai jun, *Automorphisms and Full Embeddings of Categories in Algebra and Topology*, online

infinitesimal deformations", whereas you mean "no nontrivial automorphisms". $\endgroup$almostrigid, in that it only has one autoequivalence (up to iso, I suppose) which isn't isomorphic to the identity, namely the functor $C \mapsto C^{op}$. $\endgroup$6more comments