# Space with prescribed local homology

Lets be $G_n$ sequence of abelian groups and $G_0 = \mathbb{Z}$.

Is there topological space $X$ that local homology groups at every point are those $G_n$ ? ie. $$\forall x\in X \; \forall n\in \mathbb{N}_0 \; : \: H_n(X,X\setminus \{x\}) = G_n$$

Extra question: Can $X$ be embedded in Euclidean space? Or can $X$ be compact? If no, why?

It is not hard to find space $X$ that its homology groups are $G_n$ ie $H_n(X) = G_n$.

With this space we can make space which has one point with prescribed local homology $G_n$. Let $\tilde{Y} = X \times [0,1]$. Define equivalence on $\tilde{Y}$ that $\{x,0\} \sim \{y,0\}$ for all $x,y \in X$. Now let $Y$ be factor space $Y = \tilde{Y}\big/_\sim$.

Than I think $H_n(Y,Y\setminus [\{x,0\}]) = G_n$. I do not have proof of this yet, but I think it should work.

• The example with local homology at one point indeed works and its proof is application of the long exact sequence for homology of the pair. However, getting prescribed local homology at all points looks very difficult, if possible at all, since (with few exceptions of the sequences $G_n$) such spaces cannot be CW-complexes. Apr 29 '14 at 21:36
• Maybe of help: the homology $H_*(X,A)$ of a pair of spaces $A\subset X$ is the reduced homology of the cofiber $Ci$ of the inclusion $i:A\hookrightarrow X$. May 13 '14 at 15:01
• @OlivierBégassat Thx, unfortunately I have no idea what cofiber is but I'll have a look at wiki.
– tom
May 13 '14 at 20:50

There are two subquestions in your question:

1. $H_n(Y, Y')\cong G_n$. Where $Y'= Y\setminus \{[x,0]\}$. Indeed, since $Y$ is contractible, we have the long exact sequence of reduced homology groups: $$0= \tilde{H}_k(Y)\to H_k(Y, Y') \to \tilde{H}_{k-1}(Y') \to \tilde{H}_{k-1}(Y)=0 ...$$ Since $$\tilde{H}_{k-1}(Y')\cong \tilde{H}_{k-1}(X)\cong G_{k-1},$$ this means that you almost got it right: $$\tilde{H}_n(Y,Y')\cong G_{n-1}$$ except for $n=1$, where you are off by the factor of ${\mathbb Z}$: $$\tilde{H}_1(Y, Y')\cong G_0/{\mathbb Z}.$$
2. The spaces $X$ with prescribed homology $G_n$ can be obtained as follows: You start with the $G_n$-Moore space $X_n=M(G_n,n)$, i.e., a space with $\tilde{H}_k(X_n)=0, k\ne n$ and $H_n(X_n)\cong G_n$. Then you take as $X$ the bouquet of the spaces $X_n$, $n\ge 0$.

3. Here is a sketch of a construction answering your other question (a space with prescribed local $n$-th homology $\cong G_{n-1}$ at every point). However, I did not check all the details, specifically, organizing the transfinite induction part:

Start with the space $Y$ as in Part 1.

Step 1. At every point $z\in Y$ of $Y$, except for the "tip" $y_0:= [x,0]$, attach a copy $Y_z$ of $Y$. Denote this space by $Y_1$. It is not hard to check that at every $y\in Y\subset Y_1$, we have $$H_n(Y_1, Y_1\setminus \{y\})\cong H_n(Y, Y\setminus \{y\})\cong G_{n-1}.$$ (To prove the first isomorphism use contractibility of $Y$: The spaces $Y_z$ that we are attaching are all contractible and disjoint from $y$.) Now, repeat Step 1 at every $z\in Y_1\setminus Y$ (attach a copy of $Y$ at every such $y$). Thus, we obtain $Y_2$. Next, obtain $Y_3$ by attaching copies $Y_z$ of $Y$ at all points $z\in Y_2\setminus Y_1$. Continue this process via transfinite induction of length bounded by the cardinality of $Y$. Note that parts of the transfinite induction will involve taking direct limits. Here you use the fact that homology commutes with direct limits.

• Nice, I need to go through all the details but I do not see why I need to continue up to cardinality of $Y$. It seams to me that $Y_\omega$ is sufficient. If I pick any point $x \in Y_\omega$ than it has to lie in some $Y_k \setminus Y_{k-1}$. Than my guess is that $H_n(Y_\omega,Y_\omega\setminus \{x\}) \cong H_n(Y_{k+1},Y_{k+1}\setminus \{x\})$
– tom
May 20 '14 at 7:48
• @tom: True, I was overcautious, you need only countable many steps and then take a direct limit. May 20 '14 at 15:37