1

I am here to ask for a second opinion of my understanding of how one obtains singular cohomology, in baby steps. I am specifically interested in understanding this at a basic categorical level. Below is how I would describe the process.

  1. First, we have the category of topological spaces, $\mathbf{Top}$. To $\mathbf{Top}$, we apply the singular complex functor which constructs a singular chain complex from a topological space; thus it is a functor from $\mathbf{Top}$ to $\mathbf{Ch}$, the category of chain complexes.
  2. Now, from $\mathbf{Ch}$, we get to the category of cochain complexes $\mathbf{CoCh}$ via the contravariant functor $\mathrm{Hom(-,\mathbb{R})}$ which, in the case of chain complexes obtained from the singular complex functor, gives us cochain complexes of singular cochains (dual spaces to singular chain groups.)
  3. Lastly, we consider the cohomology functor as a functor from $\mathbf{CoCh}$ to $\mathbf{Ab}$. In the case of singular cochain complexes, we obtain singular cohomology.

My questions:

  1. Was anything I said incorrect, or did anything I say stir a 'well, you should really think about that like/as ____?'
  2. In (2) and (3), I say 'in the case of (objects in a category obtained by a specific functor when applied to a specific class of objects in another category)...' Can I simply call these subcategories with no issue? The reason I ask is that I don't ever see this done.
  3. I think I can replace $\mathbb{R}$ with any abelian group, and my description of the above process would need no revision. Am I correct?

Sorry if my question is vague; I have no clear goal, except to understand everything.

Descartes Before the Horse
  • 7,386
  • 1
    This old question of mine may help you! Even though it deals with Homology, the co- prefix isn't much more work ! – Anthony Jun 08 '21 at 17:20
  • 1
    For your 2., better to understand that these constructions are functors (so you describe the cohomology functor as the composition of three functors). "Subcategories" is less relevant here. For your 3., no such thing as a module over an abelian group, but yes, can consider any abelian group. Also, maybe you want your final target category to be graded abelian groups, not just abelian groups, to keep track of the various cohomological degrees. – Sasha Jun 08 '21 at 18:02
  • 1
    Cannot edit my comment so I'll make a new one instead. I linked the wrong question! (Even though it is still relevant.) For the details of the construction, here is what I wanted to link ;) – Anthony Jun 08 '21 at 18:30

1 Answers1

2

  1. If I haven't overlooked anything, all you've said is correct. :) Judging from people's taste, your functor in 1. might be a little quick and could be factored through the singular chain functor to $\mathbf{sSet}$. It would also probably be helpful to write out diagrams/arrows to easier keep track of things and especially keep track of the contravariance.

  2. You can indeed call those the full subcategory spanned by some objects. (But note that the functors themselves are OK on the entire category.)

  3. Yes. E.g. $\mathbb{Z}$ is important.

Qi Zhu
  • 9,413