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In discussing the Homology group of CW-complexes:

Theorem: Let $X^*$ be a space obtained by appropriately attaching a collection of $n$-cells $\{e^n_\lambda | \lambda \in \Lambda\}$ to $X$. Then $H_q(X^*,X) = 0$ for all $q \neq n$. For each index $\lambda \in \Lambda$, the characteristic map $f_\lambda$ induces a monomorphism of relative homology groups $$f_{\lambda *} : H_n(E^n, S^{n-1}) \to H_n(X^*,X)$$ and $H_n(X^*,X)$ is the direct sum of the image subgroups.

Intuitively, I can understand the first part $H_q(X^*,X) = 0$ for all $q \neq n$ as follows: $H_q(X^*,X)$ is kind of like the homology group of the space $X^*$ if we 'trivialise' the subspace $X$. But if we are ignoring $X$ from $X^*$, then what we are left with are the $n$-cells $e^n_\lambda$ that we pasted onto $X$ to make $X^*$. These are all $n$-dimensional balls, so if we are considering the homology of a higher dimension $q > n$ then $H_q(X^*,X) = 0$ because nothing of higher dimension can be fit into these balls, and if we are considering the homology of a lower dimension $q < n$ then $H_q(X^*,X) = 0$ because all lower dimension 'paths' get trivialised (the way I imagine it in my head is like how the fundamental group of $S^2$ is trivial?)

Firstly, is this an acceptable 'intuitive' understanding of the first statement? And how would I go about understanding the rest of the theorem? It seems 'intuitive' that $H_n(E^n,S^{n-1})$ is just $\mathbb{Z}$ (I think) because effectively we are starting off with a closed $n$-dimensional ball, then we are identifying its boundary together, which sort of "zips it up" into an $n$-dimensional sphere, which has exactly one hole in it. I'm not sure how to go ahead and think about the characteristic map though...

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I do not think that it is a good idea to think of trivialization of paths. Homology is more about cycles and boundaries. Remember for a CW complex $(X,\{X^{n}\}_n)$ cellular chain groups are defined as $$ C_n(X) = H_n(X^{n},X^{n-1})$$ In the question we have $C_{n}(X^{\ast}) = H_n(X^{\ast}, X) \cong \mathbb{Z}^{\text{number of n-cells}}$. You are right for the case $q>n$ but it might be better to obtain the result by observing all cellular chain groups are zero for $q>n$ by definition of chain groups. For the case $q <n$, you can think of cellular approximation theorem((image of every singular $q$-chain can be homotoped to $(q+1)$-skeleton) and conclude that all $q$-cycles lies in $(q+1)$-skeleton (here recall that relative singular chains groups are defined as $S_q(X^{\ast},X)= S_q(X^{\ast})/S_q(X)$) or using the fact that $(X^{\ast},X)$ is a good pair and this gives $H_q(X^{\ast},X) \cong \tilde{H}_q(X^{\ast}/X)= \tilde{H}_q(\vee_{\lambda \in \Lambda} S^n)=0$ when $q <n$.

Your intuition about $H_n(E^n,S^{n-1})= \mathbb{Z}$ is right $(E^{n},S^{n-1})$ is a good pair and $H_n(E^{n},S^{n-1}) \cong \tilde{H}_n( E^{n}/S^{n-1}) \cong \tilde{H}_n(S^n) \cong \mathbb{Z}$.

For the characteristic maps recall that $f_{\lambda}|_{int{E^{n}}} : int(E^{n}) \to X^{\ast}$ is a homeomorphism onto its image and $f_{\lambda}|_{S^{n-1}} (S^{n-1}) \subset X$ (Image of attaching map lies in $(n-1)$-skeleton.) Hence, we can conclude that $f_{\lambda}$ is a map of pairs $(E^n ,S^{n-1}) \to (X^{\ast},X)$ and since it restricts to a homeomorphism on interior of $E^{n}$, it induces an injection(monomorphism) on homology $$f_{\lambda \ast} : H_n(E^n,S^{n-1}) \to H_n(X^{\ast},X)$$

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