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Let $(X_i,\phi_i^j)$ be a directed system of topology spaces and its direct limit is $(X,\phi_i)$ $$\lim_{\rightarrow}(X_i,\phi_i^j)=(X,\phi_i)$$ Since $H_n$ ($n^{th}\, homology \,\,group$ ) is functor so $(H_n(X_i),(\phi_i^j)_*)$ is a directed system in category of abelian groups such that $(\phi_j)_*(\phi_i^j)_*=(\phi_i)_*$ for every $i\leq j$. I know direct limit exist for any directed system in category of groups then we can assume $$\lim_{\rightarrow}(H_n(X_i),\phi_{i^j_*})=(G,f_i)$$ and by defination of direct limit there exit a unique homomorphism $$h:G\rightarrow H_n(X)$$ such that $\phi_{i_*}=h(f_i)$ for every i.
If I show $h$ is an isomorphism then $$H_n(\lim_{\rightarrow}(X_i,\phi_i^j))\cong\lim_{\rightarrow}\left(H_*(X_i),(\phi_i)_*\right)$$ Can some body help me to proving the bijection of $h$?

Mojtaba
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2 Answers2

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It is not true that $h$ is an isomorphism in general. For instance let $X=S^1$ and consider the directed system of all countable subspaces of $X$ (with their inclusion maps). Then $X$ is the direct limit of this system (since a subset of $X$ is closed iff it is sequentially closed). But $H_1(X_i)=0$ for each $i$ in the system, since each $X_i$ is totally disconnected. So $$\lim_{\rightarrow}\left(H_1(X_i),(\phi_i)_*\right)=0$$ whereas $$H_1(\lim_{\rightarrow}(X_i,\phi_i^j))=\mathbb{Z}$$ for this system.

Eric Wofsey
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  • What fails is then that $\varinjlim S(X_i)$ is different from $S(X)$, correct? – Pedro Mar 07 '17 at 06:02
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    Right, this is true if you're talking about simplicial sets instead of spaces. – Eric Wofsey Mar 07 '17 at 06:03
  • More generally, if every compact subset of $X$ is contained in some $X_i$, the claim holds. – Pedro Mar 07 '17 at 06:04
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    @PedroTamaroff: If each $X_i$ is a subspace of $X$, then yes. But in general the maps $X_i\to X$ may not be injections, so you would need to say any map from a compact subspace factors through some $X_i$. – Eric Wofsey Mar 07 '17 at 06:06
  • @EricWofsey: My directed system has been define on directed directed index set – Mojtaba Mar 07 '17 at 06:06
  • @A-M: The countable subsets of $X$ (and their inclusions) are directed, since the union of two countable subsets is countable. – Eric Wofsey Mar 07 '17 at 06:07
  • you mean homology can't commute with direct limit in general? – Mojtaba Mar 07 '17 at 06:07
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    That's correct, homology of spaces does not commute with direct limits in general. – Eric Wofsey Mar 07 '17 at 06:16
  • why $X_i$ is totally disconnected? – Mojtaba Mar 07 '17 at 07:11
  • See http://math.stackexchange.com/questions/96541/connected-metric-spaces-with-at-least-2-points-are-uncountable – Eric Wofsey Mar 07 '17 at 07:16
  • Ok , could you please explain more about the how $X$ is direct limit of its countable subspace? – Mojtaba Mar 07 '17 at 07:26
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    Sorry, the explanation is potentially quite complicated depending on what you already know, and I'm afraid I'm not up to it right now. I would suggest first trying to understand why any set $X$ is the direct limit of its countable subsets, in the category of sets. If you really want a detailed explanation, you might ask that as a new question. – Eric Wofsey Mar 07 '17 at 07:56
  • I think that is not true that any $X$ is direct limit of its countable subsets, – Mojtaba Mar 07 '17 at 08:10
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The directed system of topoloical spaces gives a directed sistem of complexes $S_i = S(X_i)$ where $S$ is the singular chains functor. However, it is not necessarily true that $\varinjlim S_i = S$. If this were the case, the following general considerations apply.

Consider any diagram $I$, and the functor $\varinjlim : A^I \to A$. This is right exact, and then there is a canonical map

$$\theta : \varinjlim H(S_i) \to H(\varinjlim S_i)$$

which may fail to be an isomorphism. However, if your indexing set $I$ is directed, then the functor $\varinjlim : A^I \to A$ is exact, and exact "functors commute with homology". More precisely, there is a map

$$\psi : H(\varinjlim S_i)\to \varinjlim H(S_i)$$

which is an inverse to $\psi$.

Pedro
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  • My indexing set I is directed. Could you please explain more about the exactness ! and what is the $A$ and why there exist a cononical map ? – Mojtaba Mar 07 '17 at 07:09