Injective homomorphism between $H_n(\mathbb{D}^n,\mathbb{S}^{n-1}) \longmapsto H_n(X^{*},X)$

Question

I started to study $CW$-complex and I've got some questions about the following theorem since the concept doesn't feel intuitive to me.

Theorem : Let $X^*$ be obtained from $X$ attaching $n-$cells $(e_{\lambda})_{\lambda \in \Lambda}$. Then $H_q(X^*,X) = 0$ if $q \ne n$. For $\lambda \in \Lambda$ $f_\lambda$ (characteristic map) induces an injective homomorphism between $H_n(\mathbb{D}^n,\mathbb{S}^{n-1}) \longmapsto H_n(X^{*},X)$ and the last one is the direct sum of the images of those homomorpshism.

Let $a_{\lambda}$ be the $0$ of the disk given by the homeomorphism $f_{\lambda}$ of the $n$-cells with $U^n$ (where $U^n = \{x \in \mathbb{R}^n : \lvert \lvert x \rvert \rvert < 1\})$, $d_{\lambda}$ be homeomorphic to disks of radiuse $0 < \varepsilon < 1$ given by the restriction of $f_{\lambda}$ and $D := \bigcup_{\lambda} d_{\lambda}, A = (a_{\lambda})_{\lambda \in \Lambda}$, where $X^* = X \bigsqcup e_\lambda^n$

During the proof some isomorphism are given, in particular I don't understand the following two :

$1.\hspace{0.2cm}f : H_q(D,D\backslash A) \longrightarrow H_q(X^*,X^*\backslash A)$. This should follow from excision, but I don't see how taking $X^* \backslash D$ and $X^* \backslash A$ the hyphotesis of excision i.e $\text{Cl}(X^* \backslash D) \subseteq \text{int}(X^* \backslash A)$ could be verified.

$2. \hspace{0.2cm} g : H_q(X^*, X) \longrightarrow H_q(X^*, X^* \backslash A)$. This should follow from the fact that $X$ is a deformation retract of $X^*\backslash A$, but I don't see why, since to me $X^*\backslash A = X \bigsqcup_{\lambda} (e_{\lambda}^n - a_{\lambda})$.

Any help in understanding the problem or solution would be appreciated.

Answer 1

1. $A$ is a set of isolated points of $X^*$, thus it is closed in $X^*$ and therefore $\text{int}(X^* \backslash A) = X^* \backslash A$. The $d_\lambda$ are closed disks and their interiors in $X^*$ are open sets $V_\lambda$ containing $a_\lambda$. Hence $X^* \backslash D \subseteq X^* \setminus \bigcup V_\lambda$. Since the latter is closed we get $\text{cl}(X^* \backslash D) \subseteq X^* \setminus \bigcup V_\lambda \subseteq X^* \setminus A$.

2. The closed cells $e^n_\lambda$ are the images of associated characteristic maps $\phi_\lambda : D^n \to X^*$. Thus $e^n_\lambda \setminus a_\lambda$ is the image of $D^n \setminus \{0\}$ under $\phi_\lambda$. But $S^{n-1}$ is a strong deformation retract of $D^n \setminus \{0\}$ and therefore $\phi_\lambda(S^{n-1}) \subseteq X$ is a strong deformation retract of $e^n_\lambda \setminus a_\lambda$ (note that the strong deformation retraction on $D^n \setminus \{0\}$ is compatible with the quotient map).