# Cech cohomology as a colimit over maps to a CW complex

Given a topological space $X$, we consider the following category $\mathsf{CW}_{X\to}$. The objects are finite CW complexes $Y$ equipped with a continuous map $X\to Y$. The morphisms are continuous maps $Y\to Y'$ such that the composition $X\to Y\to Y'$ agrees with $X\to Y'$. Now there is a functor: $$\mathsf{CW}_{X\to}\to\mathsf{AbGrp}$$ given by taking the cohomology of $Y$. I wish to consider the colimit of this functor: $$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$$

Now let's assume that $X$ is compact Hausdorff. I believe that in this case $\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$ is naturally isomorphic to the Cech cohomology of $X$. However, I am a bit unsure as to whether I should take the category $\mathsf{CW}_{X\to}$ as I have defined it above, or "soften" it by taking only homotopy classes of maps $X\to Y$ and homotopy classes of maps $Y\to Y'$ (maybe we could call this category $\mathsf{CW}_{X\to}^h$).

How should I think about the relationship betwee $\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$ and $\operatorname{colim}H^\ast(\mathsf{CW}_{X\to}^h)$?

Is one easier to deal with theoretically (e.g. by virtue of being filtered)? Are they indeed isomorphic via the natural map: $$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to}^h)\to\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$$ induced by the forgetful functor $\mathsf{CW}_{X\to}\to\mathsf{CW}_{X\to}^h$?

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I believe that the obvious map between these colimits goes in the opposite direction. –  Tom Goodwillie Feb 23 '13 at 21:03

I believe that the proof that the colimit of cohomology over (the opposite of) $\mathsf{CW}_{X\to}^h$ coincides with Cech cohomology is straightforward.

I'm not sure if this is the same as the colimit over $\mathsf{CW}_{X\to}$ . I can see that it would be if the following statement is true. Maybe someone can tell us whether it is:

True or false? If $X$ is compact Hausdorff and $Y$ is finite CW, then for any continuous map $X\to Y^I$ there is a finite complex $Z$ and a factorization $X\to Z\to Y^I$ of the map.

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A short answer for now (a longer edit or answer perhaps later or perhaps its not needed).

When you use continuous functions then your category for a compact space $X$ is virtually just the space $X$ itself. In the given context it does not buy much. But when you use homotopy classes instead then you get the shape of compact $X$ of the continuous shape theory (as in the natural model of the axiomatic shape theory). The continuous shape functor is like a Dedekind section between shape functors and continuous homotopy invariant functors (like Cech homology and cohomology, etc)--these latter ones are automatically shape invariant. Topological category of compact spaces is like Earth, and the continuous shape functor is like a cosmic station for all these other "cosmic" functors.

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