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This is a review question I'm doing for an upcoming exam. Consider $X$ a Hausdorff space, and $(H_*(X),\partial_*)$ its singular homology. I must prove that, given $[z]\in H_p(X)$ there exists a compact $C\stackrel i\hookrightarrow X$ such that $[z]$ is the image of $[y]\in H_p(C)$ by the induced map $i_p$ ($i$ originally the inclusion). Then I am to use this to prove that if $X$ has the additional property that for every compact $C\subset X$ there exists a compact $C'\subset C$ such that $H_p(C')=0$, then $H_p(X) = 0$.

I have done the first part by taking $C$ to be the reunion over $\alpha$ of $\sigma_\alpha(\Delta^p)$ for every non-zero $m_\alpha$ in the basis expansion $z=\sum_\alpha m_\alpha\sigma_\alpha$. Such a reunion will be finite and therefore compact, and defining $\hat\sigma_\alpha:\Delta^p\to C$ in the obvious way then $i\circ \hat\sigma_\alpha = \sigma_\alpha$ which implies $i_p\left[\sum_\alpha m_\alpha\hat\sigma_\alpha\right] = [z]$. I note that neither is $C$ unique, nor the element mapped to $[z]$ by $i_p$ once $C$ is chosen.

I'm lost on the final conclusion though. I have tried setting up some nice commutative diagrams with rows $H_p(C')=0\to H_p(C)\to H_p(X)$ and columns the boundary maps but I'm not able to get any conclusions out of them. Could I get some hints? Thank you.

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    $\begingroup$ The result seems weird : in general for every compact $C$ then take a point $x\in C$, and $\{x\}\subset C$ is a compact with trivial homology. Did I misunderstand something ? $\endgroup$ Apr 16, 2016 at 19:25

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You're having difficulty proving this because the statement is wrong. As stated, the condition that there exists a compact $C'\subseteq C$ such that $H^p(C')=0$ is true for every space $X$, since you can always take $C'=\emptyset$.

I would guess that the intended statement should have $C\subseteq C'$ rather than $C'\subseteq C$. In fact, with this correction, you don't even need to assume $C'$ is compact. By the first part, every class in $H^p(X)$ is in the image of $H^p(C)$ for some compact $C\subseteq X$, and then by factoring the map $H^p(C)\to H^p(X)$ through $H^p(C')$ you can conclude that the class must be trivial.

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  • $\begingroup$ I thought there was something weird about the statement. These damned misprints will be the end of my sanity. Thanks very much for your help! $\endgroup$
    – GPerez
    Apr 16, 2016 at 19:39
  • $\begingroup$ Sorry, an additional question. It's not altogether clear to me why the space is required to be Hausdorff? $\endgroup$
    – GPerez
    Apr 16, 2016 at 21:17
  • $\begingroup$ Yeah, that's totally unnecessary. Maybe the author thought it was needed in order to know that compact subsets are closed? But you don't actually need them to be closed anywhere in the argument. $\endgroup$ Apr 16, 2016 at 21:24
  • $\begingroup$ Yeah that's the typical argument with the Hausdorff condition. But as you say it's not necessary here. Oh well! $\endgroup$
    – GPerez
    Apr 16, 2016 at 21:30

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