Homology of infinite intersection

Question

If $X_1\supseteq X_2\supseteq \ldots$ is a sequence of "nice" compact spaces, I would like to know whether the natural map from $H_*(\cap X_i)$ to the inverse limit $\lim \, H_*(X_i)$ is surjective. In particular, if there exist nonzero $\beta_i\in H_q(X_i)$ such that the inclusion-induced homomorphism $i_*$ takes $\beta_i$ to $\beta_{i-1}$, is there an $\alpha\in H_q(\cap X_i)$ such that the inclusion-induced image of $\alpha$ is the $\beta_i$ for each $i$?

I have found Milnors paper "On the Steenrod homology theory" where he shows, using Steenrod homology theory, that there exists a surjective map $H_q(\cap X_i)\to \lim H_q(X_i)$. However, it is not completely clear from the text whether the preimage of $(\beta_1,\beta_2,\ldots)\in \lim H_q$ is mapped to $\beta_i$ by the inclusion-induced homomorphism $H_q(\cap X)\to H_q(X_i)$.

Answer 1

To expand on my answer above, consider the sequence $\dots\subset Y_2\subset Y_1$ of spaces where $Y_i=X_i$ for $i\leq n$ and $Y_i=X_n$ for $i>n$.

Milnors proof proceeds by taking $M_i$ to be the mapping cylinder of $X_{i+1}\rightarrow X_i$ and $M_0$ the cone over $X_0$ with vertex $t$. Then one takes the union $F=\cup M_i$ with $M_i$ glued to $M_{i+1}$ along $X_{i+1}$. Then one makes a space $T=F\cup X$ by adding $X$ `at infinity'. The upshot is that $T$ can be contracted onto $t$ by collapsing everything down to $X_1$ and then using the cone $M_0$ to get down to $t$.

Now, once this is done, Milnor sets $F_1$ to be the disjoint union of the odd $M_i$ and $F_0$ the disjoint union of the even $M_i$. Then $F_0\cap F_1$ is just the disjoint union of all the $X_i$, and $F_0\cup F_1 = F$.

The proof now follows quickly by looking at the long exact sequence i homology corresponding to the triple $(T,X\cup t,t)$ together with Mayer-Vietoris for $(F_0,F_1)$.

Now carry out the same constructions for $Y$, denoting the relevant spaces by $M_i', t', T', F',F_0',F_1'$.

The point is that there are natural maps $M_i'\rightarrow M_i$ inducing maps $T'\rightarrow T$ and $F'\rightarrow F$, and the induced maps on cohomology are evidently the ones gotten by inclusion.

Now, $Y=\cap_m Y_m = X_n$ and so the map Milne gets $H_q(X)\rightarrow \lim H_q(X_i)$ commutes with the inclusion maps $H_q(X)\rightarrow H_q(X_n)$, which gives what you want.

Answer 2

Here is a counterexample, which is probably not "nice". Let $X$ be the Warsaw-circle. Let $X_n$ be the obtained from the Warsaw-circle by thickening the limit inverval by $1/n$. The intersection of all the $X_n$'s is the Warsaw-circle, and its first homology vanishes.

Each $X_n$ is homotopy equivalent to $S^1$ and the inclusion $X_{n+1}\rightarrow X_n$ is a homotopy equivalence. Thus $\lim H_1(X_i)$ is $\mathbb{Z}$ and there cannot be a surjection.

Meta: Every $X_i$ is has the structure of a compact CW-complex. However the intersection behaves badly, we cannot arrange $X_{n+1}$ to be a subcomplex of $X_n$ in this example; otherwise the intersection would be a CW-complex, which it is not. I guess this has to be a part of the niceness condition. But then on the other hand, since all spaces are assumed to be compact, we have $X_i=X_{i+1}$ almost always and thus both inverse systems stabilize. I have no clue what a good niceness condition could be.

Answer 3

There is no reason for a homological surjection. And it's false all the time, so-to-speak. Thus let's assume that we are asking either about a homological injection or about the cohomological surjection which goes in the other direction (naturally--as a categorist would say).

Now the answer is YES under some reasonable assumptions. Consider an inverse system of Hausdorff compact spaces. They need not be metrizable, and they need not be nice. Also, the projections can be arbitrary--they need not be injective nor surjective. Just assume that the limit space $X$ is ANR (i.e. for any compact pair $(Y\ B)$, and for every continuous map $f:B\rightarrow X$ there is a neighborhood $G\subseteq Y$ of $B$, and a continuous $g:G\rightarrow X$ such that $g$ extends $f$, i.e.$f=g|B$). Then YES, when the approximating space $X_t$ has a sufficiently advanced index $t$ then we get the respective homological injection and cohomological surjection.

Indeed, we have a much stronger statement: for any advanced index $\ t\ $ the homotopy class of $\ p:X\rightarrow X_t\ $ is a homotopy $\ell$-map, i.e. it admits a continuous map $q_t:X_t\rightarrow X$ such that $\ q_t\circ p_t:X\rightarrow X\ $ is homotopic to identity. That's what it really is about.

Answer 4

For the sake of completeness, let's look at the following simple example of very nice continua, and of their inverse system of inclusions, for which there is no homological surjection.

All continua are subspaces of $\ \mathbb R^2$. Let $\ X:=(0\ 0)\ $ be a single-point space. Let $\ C(p;r)\ $ and $\ B(p;r)\ $ stand for closed and open discs which have their center in $\ p\ $ and their radius equal to $\ r.\ $ Define:

$$X_n\ \,:=\ \,C\left(\left(2^{-n}\ 0\right);\ 2^{-n}\right)\ \setminus\ B\left(\left(\frac 3{2^{n+1}}\ 0\right);\ 2^{-(n+1)}\right)$$

Thus $\ H_*(X)=0\ $ while $\ H_*(X_n) = H_*(S^1).\ $ As we see, there are no (homological) surjections.