Inverse limit of cohomology of direct system of topological spaces induces an isomorphism in cohomology?

Question

Let $X_n$ be topological spaces such that $X_n \subset X_{n+1}, \forall n \in \mathbb{N}$ and let $X = \bigcup_{i} X_i $.

The $X_i$ spaces together with the inclusions $X_i \to X_j$ for $i \leq j$, is a direct system of top. spaces and the inclusions $X_i \to X$ is direct system of mappings. The cohomology functor gives rise to an inverse system of rings $H^*(X_n)$ together with an inverse system of mappings $ H^*(X) \to H^*(X_i)$ induced by the inclusions.

The above results to an induced ring morphism from $ H^*(X) \to \varprojlim H^*(X_n)$.

I want to prove that if the morphisms $H^k(X) \to H^k(X_n)$ induced by inclusions are isomorphism for $k \leq n$, then the above ring morphism is an isomorphism.

There is already a more general question about that, this one. But, I can't follow the answer since I am not familiar with the abstraction of the colimits and I can't guess what additional knowledge beyond the definitions are needed.

I think I can show that is an injection with algebraic and set-theoretic reasoning, but I am having hard time showing that it is a surjection.

I would like a help on the surjection part but without using categorical arguments like the above question. Also, I would like to ask what does "$X_n$ is a direct system of topological spaces approximating $X$" means in the question mentioned above.

Answer 1

The result is not true. Take for example $\mathbb{RP}^{\infty}=\bigcup_{n\ge1}\mathbb{RP}^n$. This is a CW-filtration, so it satisfies the hypotheses. However, all cohomology be with $\mathbb{F}_2$-coefficients, $H^{\ast}(\mathbb{RP}^n)=\mathbb{F}_2[x]/(x^{n+1})$ is a truncated polynomial ring and the inclusion $\mathbb{RP}^n\rightarrow\mathbb{RP}^{n+1}$ induces the natural projection $\mathbb{F}_2[x]/(x^{n+2})\rightarrow\mathbb{F}_2[x]/(x^{n+1})$ on cohomology rings (this is the quotient by the ideal $(x^{n+1})/(x^{n+2})$). This means the limit $\lim H^{\ast}(\mathbb{RP}^n)=\lim\mathbb{F}_2[x]/(x^{n+1})=\mathbb{F}_2[[x]]$ is the ring of formal power series over $\mathbb{F}_2$. However, $H^{\ast}(\mathbb{RP}^{\infty})=\mathbb{F}_2[x]$ is the ring of polynomials over $\mathbb{F}_2$ and the induced map $H^{\ast}(\mathbb{RP}^{\infty})\rightarrow\lim H^{\ast}(\mathbb{RP}^n)$ corresponds to the natural inclusion $\mathbb{F}_2[x]\rightarrow\mathbb{F}_2[[x]]$, which is injective yet not an isomorphism.

So, what's going on here? The issue is that limits don't commute with direct sums. Indeed, as I've explained in the comments, your hypotheses easily imply that the natural maps $H^k(X)\rightarrow\lim H^k(X_n)$ are isomorphisms for every $k\ge0$. As abelian groups, we have $H^{\ast}(X)=\bigoplus_{k\ge0}H^k(X)$ and $H^{\ast}(X_n)=\bigoplus_{k\ge0}H^k(X_n)$ for all $n\ge0$. The direct sum of these isomorphisms is an isomorphism (of abelian groups) $H^{\ast}(X)=\bigoplus_{k\ge0}H^k(X)\rightarrow\bigoplus_{k\ge0}\lim H^k(X_n)$. There is a canonical comparison map $\bigoplus_{k\ge0}\lim H^k(X_n)\rightarrow\lim\bigoplus_{k\ge0}H^k(X_n)=\lim H^{\ast}(X_n)$ that compares the limit of direct sums with the direct sum of limits (i.e. the statement that limits and direct sums commute would be that this map be an isomorphism). The composite of these is the natural map $H^{\ast}(X)\rightarrow\lim H^{\ast}(X_n)$, so its failure to be an isomorphism is precisely the failure of limit and direct sum to commute.

Let's analyze this failure further. This is purely an algebraic issue, so let's suppose we have sequences $\dotsc\rightarrow A_{i+1,k}\rightarrow A_{ik}\rightarrow\dotsc\rightarrow A_{0k}$ of abelian groups for every $k\ge0$ (the same would work equally well with an arbitrary index set). Then, we have the canonical homomorphism $\bigoplus_{k\ge0}\lim A_{ik}\rightarrow\lim\bigoplus_{k\ge0}A_{ik}$. The RHS consists of tuples $(x_{ik})\in\prod_{i,k\ge0}A_{ik}$ such that $x_{i+1,k}$ maps to $x_{ik}$ under $A_{i+1,k}\rightarrow A_{ik}$ for every $i,k\ge0$ and such that, for every $i\ge0$, there are only finitely many $k\ge0$ such that $x_{ik}\neq0$. The LHS consists of tuples $(x_{ik})\in\prod_{i,k\ge0}A_{ik}$ such that there are only finitely many $k\ge0$ such that $x_{ik}\neq0$ for some $i\ge0$ and $x_{i+1,k}$ maps to $x_{ik}$ under $A_{i+1,k}\rightarrow A_{ik}$. Note the difference? The cardinality $|\{k\ge0\mid x_{ik}\neq0\}|$ is just finite for each $i\ge0$ in the former case, but in the latter case it is uniformly finite in $i\ge0$. In the example I gave at the start, you can see this with the infinite sum $1+x+\dotsc+x^n+\dotsc\in\mathbb{F}_2[[x]]$. Its projection onto each $\mathbb{F}_2[x]/(x^{n+1})$ is the finite sum $1+x+\dotsc+x^n$, but the number of summands increases without bound in $n$. Whilst each finite sum $1+x+\dotsc+x^n$ can be interpreted as an element in $\mathbb{F}_2[x]$, the infinite sum $1+x+\dotsc+x^n+\dotsc$ cannot.

However, the discussion just now implies at least that the comparison map is always injective, so $H^{\ast}(X)\rightarrow\lim H^{\ast}(X_n)$ does exhibit the LHS as a subring of the RHS.

Answer 2

Lets start with some basics. $X_n$ is a direct system of topological spaces approximating $X$ means exactly that $X$ is the direct limit of $X_n$ (or at least this is the implied meaning) so $X = \varinjlim X_i$ or as you noted, $X = \bigcup_i X_i$. Now, you have an inverse system of rings and a map from $H^*(X)$ to the inverse system so that everything commutes, so by the universal property of inverse limits, you have a map $H^*(X) \to \varprojlim H^*(X_i)$. I haven't really stated anything new so far.

Let's try doing this in two steps. First, let's do this at the level of a specific cohomology group. So now let's fix $k$. We have a map $ H^k(X) \to \varprojlim H^k(X_i)$ for the same reason as above. What does this look like? An element of the inverse limit looks like a sequence $(a_0,...)$ such that $a_i$ is the image of $a_{i+1}$ under the induced morphism $H^k(X_{i+1}) \to H^k(X_i)$. We know for some $n$, we have $H^k(X) \cong H^k(X_n)$ for all $k < n$. So let's cut these sequences short at $n$ and instead only consider the subset of sequences that look like $(a_0,...,a_n)$. See if you can figure out why this truncated sequence is still isomorphic to the infinite sequence inverse limit (we don't lose any information by cutting off everything after $n$ since everything is isomorphic beyond that point, so just use the universal property of the inverse limit and show these objects are isomorphic). Great, so now I claim $H^k(X) \cong \varprojlim H^k(X_i)$. What are the maps? Well, we send $a_i \in H^k(X)$ to its corresponding element in $H^k(X_n)$ (let's call this $\alpha_n$) and then look at the element $(\alpha_0,.....,\alpha_n)$. What's the reverse map? Take $\alpha_n$ and map it to $a_n$ using the isomorphism. Great! So degree wise, this isomorphism holds in each cohomology group because there is some stabilization. This is expected, and in fact the question you link to says they could see this would hold degreewise.

See if you can figure out how to show that this isomorphisms in each degree will piece together to form a ring isomorphism!