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I am given $A,B\subseteq X$ such that $Int A\cup Int B=X$. I need to prove that inclusion $(X,A\cap B)\hookrightarrow (X,A)$ induces epimporhism in realtive homology groups. Is this a right argument:

From the naturality of long exact homology sequences we have that the following diagram commutes

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{} H_p(X) & \ra{\pi_1} & H_p(X,A\cap B) \\ \da{i_1} & & \da{i_2} \\ H_p(X) & \ra{\pi_2} & H_p(X,A), \end{array} $$ where $\pi_1$ and $\pi_2$ are projections and $i_1$ and $i_2$ are inclusions. Since, $i_1$ and $\pi_2$ are epimorphisms it follows that the $i_2$ is also epimophim.

I don't need that $Int A\cup Int B=X$?

Madara Uchiha
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  • Why would $\pi_2$ be onto? – JHF Sep 01 '16 at 21:52
  • Now I see it doesn't need to be onto... – Madara Uchiha Sep 01 '16 at 22:11
  • You could try the relative Mayer-Vietoris sequence $$ \dots \longrightarrow H_{n+1}(X,Y) \longrightarrow H_n(A\cap B,C\cap D) \longrightarrow H_n(A,C)\oplus H_n(B,D) \longrightarrow \dots $$ which requires $X$ to be the union of the interiors of its subspaces $A,B$, and $Y$ to be the union of the interiors of $C \subseteq A$ and $D \subseteq B$. Now let $A = B = X$ and $C=A, D=B$. – Stefan Hamcke Sep 05 '16 at 13:52
  • Using that sequence i get that $H_p(X,A\cap B)$ and $H_p(X,A)\oplus H_p(X,B)$ are isomorphic for every $p$. Maybe i can conclude from there that my inclusion induces epimorphism but i don't know how map between $H_p(X,A\cap B)$ and $H_p(X,A)\oplus H_p(X,B)$ in obove Mayer-Vietoris sequnce looks. – Madara Uchiha Sep 05 '16 at 17:46

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