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Thm. 3.35, pg245. The proof begins 247-248.

Hatcher defines a duality map $$D:H^k_c(M;R) \rightarrow H_{n-k}(M;R) $$ where the LHS is compactly supported cohomology. An element can be represented by $\varphi \in H^k(M,M-K:R)$, $K \subseteq M$ compact.

The map is given by $D_M [\varphi] = \varphi[M|_K]$, where $M|_K$ is fundamental class of $H_n(M,M-K;R)$.

In line 10 p.248, Hatcher claims that this map, when $M=\Bbb R^n\simeq Int \Delta^n$ , can be identified with $$D_M:H^k(\Delta^n, \partial \Delta^n) \rightarrow H_{n-k}(\Delta^n)$$

How so?

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    $\begingroup$ So first of all the RHS identification should be clear due to $\mathbb{R}^n=Int\Delta^n$ and both being contractible. So what is Hatcher saying is that $H^k_c(\mathbb{R}^n)$ can be identified with $H^k(\Delta^n, \partial\Delta^n)$. This is an algebraic analogy to the geometric fact that every compactly supported function on $\mathbb{R}^n$ can be seen as a function on $\Delta^n$ that is $0$ on (some open neighbourhood of) $\partial\Delta^n$. For details I think you need to dive into definitions. I'll think about it tomorrow and get back to you hopefuly. $\endgroup$
    – freakish
    Jun 10, 2019 at 21:44
  • $\begingroup$ The above comment explains it in a nice intuition. The following can be proved via collar neighborhoods(see Proposition 3.42): If $M$ is a compact manifold with boundary, then $H^k_c(M-\partial M) \cong H^k(M,\partial M)$. (This is easy, since every compact set $K$ in $M- \partial M$ is contained in some $M- \partial M \times (0,\varepsilon)$. See also the definition of $H^k_c(X)$ in terms of direct limits, which is explained below Proposition 3.33.) Your question is the special case $M=\Delta^n$. $\endgroup$
    – blancket
    Jan 9, 2020 at 9:13

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