Let $\mathsf{CRing}\_{\mathsf{red}}$ denote the category of reduced commutative rings, and $\mathsf{Sch}\_{\mathsf{red}}$ the category of reduced schemes. Let $L : [\mathsf{CRing}\_{\mathsf{red}},\mathsf{Set}] \to [\mathsf{CRing},\mathsf{Set}]$ be the left Kan extension (for a sufficiently large version of $\mathsf{Set}$ on the right so that it exists), given by $L(X)(A) = \mathrm{colim}\_{R \to A,\, R \text{ reduced}} \, X(R)$. We have a diagram of fully faithful functors:


$$\begin{array}{c} \mathsf{Sch}\_{\mathsf{red}} & \rightarrow & \mathsf{Sch} \\ \downarrow && \downarrow \\ [\mathsf{CRing}\_{\mathsf{red}},\mathsf{Set}] & \rightarrow & [\mathsf{CRing},\mathsf{Set}]. \end{array}$$


Does it commute? In other words, if $X$ is a reduced scheme and $A$ is an arbitrary commutative ring $A$, is the canonical map
$$\mathrm{colim}\_{R \to A,\, R \text{ reduced}} \, X(R) \longrightarrow X(A)$$
bijective? This is clear when $X$ is affine. It also holds when $A$ is local.



Functorial representation of reduced schemes
This is true. Recall the following lemma:


**Lemma.** *Let $X \stackrel i\hookleftarrow Z \stackrel f\to Y$ be a span of affine schemes, where $i$ is a closed immersion. Then the pushout $P = X \underset Z\amalg Y$ in $\mathbf{Sch}$ exists and is affine. If $X$ and $Y$ are reduced, then so is $P$.*


*Proof.* If $X = \operatorname{Spec} A$, $Y = \operatorname{Spec} B$, and $Z = \operatorname{Spec} C$, then [Stacks, Tag [0ET0](https://stacks.math.columbia.edu/tag/0ET0)] shows that the pushout is represented by the affine scheme
$$P = \operatorname{Spec} \big(A \underset C\times B\big).$$
The final statement follows since products and subrings of reduced rings are reduced. $\square$


Thus, it suffices to show the following:


**Proposition.** *Let $f \colon Z \to S$ be a morphism from an affine scheme $Z$ to a reduced scheme $S$. Then there exists a reduced affine scheme $X$, a closed immersion $i \colon Z \hookrightarrow X$, and a morphism $g \colon X \to S$ such that $f = gi$.*


Indeed, this immediately settles surjectivity. Moreover, if $Z \hookrightarrow X \to S$ is a factorisation as in the proposition, and $Z \to Y \to S$ is any factorisation with $Y$ affine reduced, then $P = X \amalg\_Z Y$ is affine and reduced by the lemma above, giving a third factorisation $Z \to P \to S$ of which both $Z \hookrightarrow X \to S$ and $Z \to Y \to S$ are refinements. The explicit description of colimits in $\mathbf{Set}$ (see e.g. [Tag [002U](https://stacks.math.columbia.edu/tag/002U)]) then gives injectivity.


*Proof of Proposition.* Write $Z = \operatorname{Spec} A$. If the image of $f$ lands in an affine open $V \subseteq S$, say $V = \operatorname{Spec} B$, then we conclude by factoring $B \to A$ as $B \to B[A] \twoheadrightarrow A$, where $B[A]$ is a polynomial algebra with generators $a \in A$ (which is reduced since $B$ is). In general, cover $S$ by affine opens $V\_j$ for $j \in J$, and choose a finite standard affine open cover $\coprod\_{i=1}^n U\_i \twoheadrightarrow Z$ refining the open cover $\coprod\_{j \in J}f^{-1}(V\_j) \twoheadrightarrow Z$; say $U\_i = \operatorname{Spec} A\_{f\_i}$. Then we know that the result holds for each $U\_i$, so pick $U\_i \hookrightarrow X\_i \to S$ with $X\_i = \operatorname{Spec} R\_i$ affine and reduced and $U\_i \hookrightarrow X\_i$ a closed immersion. Consider the diagram
$$\coprod\_{i \neq j} U\_i \cap U\_j \rightrightarrows \coprod\_i X\_i.\tag{1}\label{1}$$
We claim that the colimit of \eqref{1} exists in $\mathbf{Sch}$ as an affine reduced scheme. Indeed, for each $j \in \{0,\ldots,n\}$, consider the diagram
$$\coprod\_{i \neq j} U\_i \cap U\_j \rightrightarrows \Big( \coprod\_{i=1}^j X\_i \amalg \coprod\_{i=j+1}^n U\_i \Big)\tag{2}\label{2}.$$
We will prove by induction on $j$ that this diagram has an affine colimit in $\mathbf{Sch}$. For $j = 0$, the colimit is simply $Z$. If the diagram \eqref{2} has an affine colimit $P\_j$ for some value of $j$, then the diagram for $j+1$ has affine colimit $P\_{j+1} = P\_j \underset{U\_{j+1}}\amalg X\_{j+1}$ by the lemma above.


This proves that \eqref{1} has an affine colimit $X = P\_n$ in $\mathbf{Sch}$; say $X = \operatorname{Spec} R$. Applying the universal property to $\operatorname{Hom}(-,\mathbf A^1\_{\mathbf Z})$ shows that
$$R = \lim \Big( \prod\_{i=1}^n R\_i \rightrightarrows \prod\_{i \neq j} A\_{f\_if\_j} \Big),$$
which is reduced since each $R\_i$ is. The map $P\_0 \to P\_n$ is a closed immersion $i \colon Z \hookrightarrow X$, and the universal property of the colimit of \eqref{1} gives a morphism $g \colon X \to S$ such that $f = gi$. $\square$


