Please what is the difference between these two excision property:

Let $X$ a topological space, $A$ a sub-space of $X$ and $U\subset A$ such that $\overline{U}\subset \stackrel{\circ}{A}$ . The inclusion of the pairs $(X-U,A-U)\rightarrow (X,A)$ induced an isomorphism for all $n\in \mathbb{N},$ $$H_n(X-U,A-U)\simeq H_n(X,A).$$

and :

If $A,B\subset X$ such that $X= \stackrel{\circ}{A}\cup \stackrel{\circ}{B}$ then the inclusion $e\colon (A,A\cap B)\rightarrow (X,A)$ induced an isomorphism $e_*\colon H_k(A,A\cap B)\rightarrow H_k(X,A), \forall k\in \mathbb{N}.$

  • $\begingroup$ The second isomorphism isn't $H_k(B,A\cap B)\to H_k(X,A)$? $\endgroup$ – Hamou Aug 22 '14 at 17:52
  • $\begingroup$ @Hamou i posted an answer what do you think? in the book it is $H_n(A,A\cap B)$ $\endgroup$ – Vrouvrou Aug 22 '14 at 17:59
  • $\begingroup$ Mais il faut changer le deuxième isomorphisme dans l'énoncé. $H_k(B,A\cap B)$ a la place de $H_k(A,A\cap B)$ $\endgroup$ – Hamou Aug 22 '14 at 18:01
  • $\begingroup$ Oui ou $B=X-U$ mais l'inclusion reste juste parce que on a pas d’hypothèses sur $A$ et $B$ juste que $X=\overset{º}{A}\cup \overset{º}{B}$ $\endgroup$ – Vrouvrou Aug 22 '14 at 18:05
  • $\begingroup$ Mais comparer avec le résultat que vous avez démontré! ce n'est pas la même chose. $\endgroup$ – Hamou Aug 22 '14 at 18:08

See the Theorem $1$ and $2$ here, where there is equivalent.


From the first definition we have that $X=\stackrel{\circ}{X-U}\cup \stackrel{\circ}{A}$ because $\overline{U}\subset \stackrel{\circ}{A}$ and $\stackrel{\circ}{X-U}=X-\overline{U}$

then if i put $B=X-U$ i obtain that $H_n(B,B\cap A)=H_n(X-U,A-U)$

where $A$ is the same and $X-U=B$ in this case $X=\stackrel{\circ}{A}\cup \stackrel{\circ}{B}$

so the two definitions are equivalent

since $(B,A\cap B)\rightarrow (X,A)$ still defined an inclusion so there is no contradiction

Edit: For the second implication we choose $U=X-B$ in this way as $X=\stackrel{\circ}{A}\cup \stackrel{\circ}{B}$ we have that $\overline{U}\subset A.$ and then we have the result.

  • $\begingroup$ C'est bon. a condition de corrigé l'énoncé! ce que vous avez demontrer que le premier énoncé implique le dexième. $\endgroup$ – Hamou Aug 22 '14 at 18:03

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