this end, we use the long exact sequence $$\begin{aligned}
\label{eqnles1}
\dots \rightarrow H_n(\partial M \cup {\Sigmait}, \partial M) \rightarrow H_{n}(M,\partial M)\rightarrow H_{n}(M, \partial M \cup {\Sigmait})\rightarrow H_{n-1}(\partial M \cup {\Sigmait}, \partial M)\rightarrow \dots\end{aligned}$$ in homology associated to the triple $\partial M \subset \partial M \cup {\Sigmait}\subset M$ and compute the relevant terms occurring in it by means of a collar $U \cong {\Sigmait}\times (-1,1)$ in $M$ (the collar exists by [@Lau00 Lemma 2.1.6]; see figure \[Fig:Sketch Composition\] for a pictorial presentation):

-   For the computation of $H_*(\partial M \cup {\Sigmait}, \partial M)$ we define $V:= \partial M \cap U$ and find by excision of the complement $W:= \partial M \setminus V$ of $V$ in $\partial M$ $$\begin{aligned}
        H_*(\partial M \cup {\Sigmait}, \partial M) \cong H_*((\partial M \cup {\Sigmait}) \setminus W, \partial M \setminus W). 
        \end{aligned}$$ Since the inclusion $(\Sigmait,\partial {\Sigmait}) {\longrightarrow}((\partial M \cup {\Sigmait}) \setminus W, \partial M \setminus W)$ is a homotopy equivalence we arrive at $$\begin{aligned}
     H_*(\partial M \cup {\Sigmait}, \partial M) \cong H_*({\Sigmait},\partial {\Sigmait}).\label{eqngluing1}\end{aligned}$$

-   For the computation of $H_*(M,\partial M \cup {\Sigmait})$ we use that the inclusion $(M,\partial M \cup {\Sigmait}) {\longrightarrow}(M,\partial M \cup U)$ is a homotopy equivalence. After excising ${\Sigmait}$ in $(M,\partial M \cup 