Property of excision of Homology

Question

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 $e_*\colon H_k(A,A\cap B)\rightarrow H_k(X,A), \forall k\in \mathbb{N}.$

Answer 1

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.

Answer 2

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