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A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\mathrm{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed,  let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Strong specializations (edited after Martin's comments):
Say a point $x\in X$ is a strong specialization of a point $y$ if there is a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local,  irreducible and one-dimensional).
This notion is categorical: to see this, it remains to distinguish $b$ from $a$ categorically on a scheme $T$ as above. We may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\mathrm {Spec}\,(K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\mathrm{Spec}\,(R/t^n)\to\mathrm{Spec}\,R$ ($n\geq1$) are distinct monomorphisms with image $a$.)

As Martin points out, all specializations are strong on a locally noetherian scheme, but probably not in general.

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\operatorname{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed,  let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Strong specializations (edited after Martin's comments):
Say a point $x\in X$ is a strong specialization of a point $y$ if there is a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local,  irreducible and one-dimensional).
This notion is categorical: to see this, it remains to distinguish $b$ from $a$ categorically on a scheme $T$ as above. We may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\operatorname{Spec}(K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\operatorname{Spec} (R/t^n)\to\operatorname{Spec}\ (n\geq1)$ are distinct monomorphisms with image $a$.)

As Martin points out, all specializations are strong on a locally noetherian scheme, but probably not in general.

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\mathrm{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed,  let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Immediate specializations:
Say a point $x\in X$ is an immediate specialization of a point $y$ if it is a specialization and there is no intermediate specialization, in the obvious sense. This property is categorical. To see this, note that it is equivalent to the existence of a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local,  irreducible and one-dimensional), plus the condition that there is no intermediate point in the same sense.  It remains to distinguish $b$ from $a$ categorically: we may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\mathrm {Spec}  (K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\mathrm{Spec}\,(R/t^n)\to\mathrm{Spec}\,R$ ($n\geq1$) are distinct monomorphisms with image $a$.)

Consequence: "$X$ is irreducible of dimension $\leq d$" is categorical, because it is equivalent to "$X$ has a point $y$ such that every point $x$ is obtained from $ y$ by at most $d$ immediate specializations".

Similarly, "$X$ is local of dimension $\leq d$" is categorical: replace "specialization" by "generalization" in the previous argument. As a consequence, on a finite-dimensional scheme, the localization at a point is categorical.

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\mathrm{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed,  let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Strong specializations (edited after Martin's comments):
Say a point $x\in X$ is a strong specialization of a point $y$ if there is a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local,  irreducible and one-dimensional).
This notion is categorical: to see this, it remains to distinguish $b$ from $a$ categorically on a scheme $T$ as above. We may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\mathrm {Spec}\,(K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\mathrm{Spec}\,(R/t^n)\to\mathrm{Spec}\,R$ ($n\geq1$) are distinct monomorphisms with image $a$.)

As Martin points out, all specializations are strong on a locally noetherian scheme, but probably not in general.

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\mathrm{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed,  let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Specialization of points on a scheme $X$  is categorical. To see this, note that every connected two-point scheme $T$ is local, irreducible and one-dimensional. To distinguish between the closed point $a$ and the generic point $b$, we may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. This implies  the claim: $x\in X$ is a specialization of $y$ iff there is a morphism from some $T$ as above to $X$ sending $a$ to $x$ and $b$ to $y$.

Consequence: dimension is categorical.

Local schemes are categorical: $X$ is local with closed point $x$ iff every point is a generalization of $x$. As a consequence, localizing a scheme at a point is categorical by the obvious universal property.

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\mathrm{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed,  let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Immediate specializations:
Say a point $x\in X$ is an immediate specialization of a point $y$ if it is a specialization and there is no intermediate specialization, in the obvious sense. This property is categorical. To see this, note that it is equivalent to the existence of a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local,  irreducible and one-dimensional), plus the condition that there is no intermediate point in the same sense.  It remains to distinguish $b$ from $a$ categorically: we may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\mathrm {Spec}  (K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\mathrm{Spec}\,(R/t^n)\to\mathrm{Spec}\,R$ ($n\geq1$) are distinct monomorphisms with image $a$.)

Consequence: "$X$ is irreducible of dimension $\leq d$" is categorical, because it is equivalent to "$X$ has a point $y$ such that every point $x$ is obtained from $ y$ by at most $d$ immediate specializations".

Similarly, "$X$ is local of dimension $\leq d$" is categorical: replace "specialization" by "generalization" in the previous argument. As a consequence, on a finite-dimensional scheme, the localization at a point is categorical.