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user43198
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Let $\mathcal{F}:\mathrm{Sch}/S \to \{\mathrm{Sets}\}$ be a representable functor, locally of finite presentation i.e., for any inverse limit of affine schemes $\varprojlim\limits_i Z_i$, the canonical morphism $\phi_1:\varinjlim\limits_i \mathcal{F} (Z_i)\to \mathcal{F}(\varprojlim\limits_i Z_i)$ is bijective. Denote by $X$ the scheme representing $\mathcal{F}$. The question is whether the natural tranformation $\mathcal{F}(-) \to \mathrm{Hom}_S(-,X)$ commute with inverse limits? More precisely, consider an inverse system of affine schemes $\{Z_i\}_{i \in I}$ with limit $Z:=\varprojlim\limits_i Z_i$. Consider a sequence of objects $(a_i)$$(a_i)_{i \in I}$ with $a_i \in \mathcal{F}(Z_i)$ and $f_i \in \mathrm{Hom}_S(Z_i,X)$ corresponding to each $a_i$ (under the natural transformation). Suppose that the sequence $(f_i)$ (resp. $(a_i)$) converge to some $f \in \mathrm{Hom}_S(\varprojlim\limits_i Z_i,X)$ (resp. $a \in \mathcal{F}(\varprojlim\limits_i Z_i)$). Is it true that $a$ must map to $f$ under the above natural transformation?

I do not know if this question is obvious, in which case a reference will be sufficient.

Let $\mathcal{F}:\mathrm{Sch}/S \to \{\mathrm{Sets}\}$ be a representable functor, locally of finite presentation i.e., for any inverse limit of affine schemes $\varprojlim\limits_i Z_i$, the canonical morphism $\phi_1:\varinjlim\limits_i \mathcal{F} (Z_i)\to \mathcal{F}(\varprojlim\limits_i Z_i)$ is bijective. Denote by $X$ the scheme representing $\mathcal{F}$. The question is whether the natural tranformation $\mathcal{F}(-) \to \mathrm{Hom}_S(-,X)$ commute with inverse limits? More precisely, consider a sequence of objects $(a_i)$ with $a_i \in \mathcal{F}(Z_i)$ and $f_i \in \mathrm{Hom}_S(Z_i,X)$ corresponding to each $a_i$ (under the natural transformation). Suppose that the sequence $(f_i)$ (resp. $(a_i)$) converge to some $f \in \mathrm{Hom}_S(\varprojlim\limits_i Z_i,X)$ (resp. $a \in \mathcal{F}(\varprojlim\limits_i Z_i)$). Is it true that $a$ must map to $f$ under the above natural transformation?

I do not know if this question is obvious, in which case a reference will be sufficient.

Let $\mathcal{F}:\mathrm{Sch}/S \to \{\mathrm{Sets}\}$ be a representable functor. Denote by $X$ the scheme representing $\mathcal{F}$. The question is whether the natural tranformation $\mathcal{F}(-) \to \mathrm{Hom}_S(-,X)$ commute with inverse limits? More precisely, consider an inverse system of affine schemes $\{Z_i\}_{i \in I}$ with limit $Z:=\varprojlim\limits_i Z_i$. Consider a sequence of objects $(a_i)_{i \in I}$ with $a_i \in \mathcal{F}(Z_i)$ and $f_i \in \mathrm{Hom}_S(Z_i,X)$ corresponding to each $a_i$ (under the natural transformation). Suppose that the sequence $(f_i)$ (resp. $(a_i)$) converge to some $f \in \mathrm{Hom}_S(\varprojlim\limits_i Z_i,X)$ (resp. $a \in \mathcal{F}(\varprojlim\limits_i Z_i)$). Is it true that $a$ must map to $f$ under the above natural transformation?

I do not know if this question is obvious, in which case a reference will be sufficient.

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user43198
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Representable functors and direct limits

Let $\mathcal{F}:\mathrm{Sch}/S \to \{\mathrm{Sets}\}$ be a representable functor, locally of finite presentation i.e., for any inverse limit of affine schemes $\varprojlim\limits_i Z_i$, the canonical morphism $\phi_1:\varinjlim\limits_i \mathcal{F} (Z_i)\to \mathcal{F}(\varprojlim\limits_i Z_i)$ is bijective. Denote by $X$ the scheme representing $\mathcal{F}$. The question is whether the natural tranformation $\mathcal{F}(-) \to \mathrm{Hom}_S(-,X)$ commute with inverse limits? More precisely, consider a sequence of objects $(a_i)$ with $a_i \in \mathcal{F}(Z_i)$ and $f_i \in \mathrm{Hom}_S(Z_i,X)$ corresponding to each $a_i$ (under the natural transformation). Suppose that the sequence $(f_i)$ (resp. $(a_i)$) converge to some $f \in \mathrm{Hom}_S(\varprojlim\limits_i Z_i,X)$ (resp. $a \in \mathcal{F}(\varprojlim\limits_i Z_i)$). Is it true that $a$ must map to $f$ under the above natural transformation?

I do not know if this question is obvious, in which case a reference will be sufficient.