Representable functors and direct limits

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.

• I think you got some 'variance' issues: what do you mean by $(f_i)$ converges to $(f)$? The natural map is $\varinjlim_i \mathrm{Hom}_S(Z_i,X)\to \mathrm{Hom}_S(\varprojlim_i Z_i,X)$, so every single $f_i$ will determine $f$. Another comment is that (in case you're assuming that) it's not always true that the above map is an isomorphism... It should be compatible with the natural isomorphism $F\cong \mathrm{Hom}_S(-,X)$, though. – Mattia Talpo Oct 19 '15 at 17:24
• You seem to be convinced that elements of a direct limit are collections of compatible elements, but that's what you get with inverse limits. The direct limit is rather a quotient of the union of the sets, and because of this, every element in any of the sets gives you an element of the direct limit, and every one of these arises this way. Anyway, Theo already answered below. – Mattia Talpo Oct 20 '15 at 17:31
• @MattiaTalpo Sorry, my mistake. I have removed my comment. – user43198 Oct 21 '15 at 18:07

(1) "[T]he natural transformation $\mathcal F(-) \to \mathrm{Hom}_S(-,X)$" is an isomorphism. This is what it means when you say "$X$ represents $\mathcal F$". Isomorphism commute with everything categorical.
(2) Normally we think about functors commuting with limits, not natural transformations. If $\mathcal F$ is any contravariant functor and $\{X_i\}_{i\in I}$ is any diagram, then there is a canonical comparison map $$\varinjlim_i \mathcal F(X_i) \to \mathcal F(\varprojlim_i X_i).$$ $\mathcal F$ "takes limits to colimits" if this comparison map is an isomorphism for every diagram.
Suppose that $\mathcal F$ and $\mathcal G$ are two contravariant functors and that $\eta : \mathcal F \to \mathcal G$ is a natural transformation. Given a diagram $\{X_i\}_{i\in I}$, naturality of $\eta$ says that the following diagram commutes: $$\begin{matrix} \mathcal F(\varprojlim X_i) & \overset{\eta(\varprojlim X_i)}\longrightarrow & \mathcal G(\varprojlim X_i) \\ \uparrow & & \uparrow \\ \varinjlim \mathcal F(X_i) & \overset{\varinjlim \eta(X_i)}\longrightarrow & \varinjlim \mathcal G(X_i)\end{matrix}$$ If both $\mathcal F$ and $\mathcal G$ take limits to colimits, then the vertical arrows are isos. If $\eta$ is an isomorphism, then the horizontal arrows are isos. Regardless, this is perhaps a statement of the form "natural transformations commute with (co)limits."