Does this functor commute with inverse limits? Let $F$ a contravariant functor from the category $A$ of pointed connected CW-complexes up to homotopy to the category $B$ of pointed sets, with $F$ sending coproducts of $A$ to products of $B$. Let $X_i$ be an ascending series of objects in $A$. Is it true that $F\left(\underset{\longleftarrow}{\lim}X_i\right) = \underset{\longleftarrow}{\lim}F(X_i)$, where inverse limit is defined here?
To be more precise I am trying to understand why the following application is defined $F(f) : F(X) \to \underset{\longleftarrow}{\lim}F(X_i)$, with $X= \bigcup_{i}X_i$. I was presuming that $f : \underset{\longleftarrow}{\lim}X_i \to X_i \to X$ and, given that $F$ reverses all the arrows, I would get $F(X) \to F\left(\underset{\longleftarrow}{\lim}X_i\right)$, but I am not totally sure $F$ commutes with the inverse limit. Is there a different reason why $F(f)$ is defined?
 A: A contravariant functor can -if anything- send direct limits to inverse limits (this is due to the fact that direct = inverse in the opposite category) or dually inverse limits to direct limits; of course, not every contravariant functor has this property.
Since the terminology "direct" and "inverse" limit is a little bit outdated, let me speak about limits (=inverse, or projective $\varprojlim$) and colimits (=direct, or inductive, $\varinjlim$). Contravariant representable functors send colimits to limits, in the sense that $F(\text{colim})\cong \lim F$, and in general they do not transform limits into colimits, apart in very special cases.
Now, if your spaces $X_i$ are organised as a chain of inclusions
$$ X_0 \subset X_1 \subset X_2 \subset \cdots \subset X_n \subset \cdots X=\text{colim } X_i$$
by applying the contravariant $F$ to this chain one obtains a "opchain"
$$ FX_0 \supset FX_1 \supset FX_2 \supset \cdots \supset FX_n \supset \cdots FX=F\big(\text{colim } X_i\big)$$
The universal property of the limit $\lim X_i$, now, yields a unique map $c : F\big(\text{colim } X_i\big) \to \lim FX_i$. So far, no topology was involved (in the sense that this construction carries over to every category and every functor, provided the co/limits exist), but of course, when this construction is specialised to the/a category of topological spaces, there is a canonical topology on the colimit $X$ in question.
Is it the case that you are interested in this map, and not in a putative $f : \underset{\longleftarrow}{\lim}X_i \to X_i \to X$? It is a very natural question, now, to ask whether $c$ is a bijection, and there are theorems in that direction: as I mentioned, if $F$ is representable, it has this property.
If this is not the case, and you're sure that $f$ is what you want, I will show you that $f$ turns out to be a complicated way to express the CW triple $(X_0, X_i,X)$:

*

*first, observe that the object $\varprojlim X_i$ is just homeomorphic to the head of the chain, $X_0$, by virtue of the fact that the limit of a diagram indexed over a category having an initial object coincides with evaluation on said initial object. The chain of spaces in question can be interpreted as a functor from the partially ordered set $\mathbb N = \{0\le 1 \le 2 \le \cdots\}$, and in such category $0$ is an initial object.

*Given this, there is a canonical choice of a map $X_0 \to X_i$ for $i\ge 0$, the inclusion $X_0 \subseteq X_i$ that composes the first $i$ steps of the chain.

*Given this, there is also a canonical choice of map $X_i \to X$, since $X$ is the colimit of the chain. Of course, such colimit just amounts to the set-theoretic union $\bigcup X_i$ endowed with the initial topology with respect to the subspaces $X_i$. (This is as far as I would go to say that "there is a topological reason" that defines $f$)

*All in all, hence, $f$ is the inclusion of $X_0$ in $X_i$, followed by the inclusion of $X_i$ in $X$.

