Condition on compact subsets implying null singular homology

Question

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.

Answer 1

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.