Let $(Y, y_0)$ and $(Y’, y_0)$ pointed CW-complexes, with $Y’$ obtained from $Y$ by attaching $n+1$-cells. Why is it true that $i_{*}: \pi_{q}(Y, y_0) \to \pi_{q}(Y’, y_0)$ is an isomorphism for $q < n$ and an epimorphism for $q=n$? I am aware of the fact that this is true if $Y$ is the $n$-skeleton of $Y’$, but here I only know that the $n$-skeleton of $Y’$ is contained in $Y$. By the way, this has already been asked here, but I couldn’t quite understand the answer.
As a related question: if $e_{\alpha}^{n+1}$ is a $n+1$-cell in $Y’$, why is it true that $i_{*}(\alpha) =0$?
Note I am using notation as in Switzer, Lemma 9.8 in the chapter dedicated to Brown theorem.
By cellular approximation, up to homotopy, one may assume that $Y'$ is obtained by adding $n+1$-cells to the $n$-skeleton of $Y$, so that $Y^{(n)}= Y'^{(n)}$
Let $f:S^q\to Y'$ be a pointed map, $q\leq n$. Then by cellular approximation, up to pointed homotopy, $f$ lands in the $n$-skeleton of $Y'$. But the $n$-skeleton of $Y'$ is the same as that of $Y$, so $f$ lands in $Y$ up to homotopy. This shows surjectivity for $q\leq n$.
Similarly, if $q<n$, because then $q+1\leq n$, again by cellular approximation, any homotopy between two such maps can be homotoped to live in $Y$, and so this shows injectivity for $q<n$.
For the second question, if you've added a cell along $\alpha: S^n\to Y$, then that means, by definition, that there is a commutative diagram :
$\require{AMScd}\begin{CD} S^n @>>> Y\\ @VVV @VVV \\ D^{n+1} @>>> Y'\end{CD}$
it follows that $S^n\to Y\to Y'$ extends to $D^{n+1}$, i.e. it is $0$ in homotopy.