At page 136 of Hatcher's book he says:
By excision, the central term $H_n(S^n, S^n - f^{-1}(y))$ in the preceding diagram is the direct sum of the groups $H_n(U_i, U_i-x_i) \approx \mathbb{Z}$.
Here $n>0$, $f:S^n \longrightarrow S^n$ is a continuous function which $y \in S^n$ has finite pre-image $x_1,...,x_m$ and $U_i$ are disjoint open subsets of $S_n$ containing $x_i$.
I understand why it happens, and then we have $H_n(S^n, S^n - f^{-1}(y)) \approx \mathbb{Z}\oplus \ldots\oplus \mathbb{Z}$ (m-times). But we know that $H_n(S_n) \approx \mathbb{Z}$ and we see $H_n(S^n - f^{-1}(y))$ as a subgroup of $H_n(S_n)$ to define the quotient $H_n(S^n, S^n - f^{-1}(y)) \dot{=} \frac{H_n(S^n)}{H_n(S^n - f^{-1}(y))}$. So the isomorphism $\xi : H_n(S^n) \longrightarrow \mathbb{Z}$ sends $H_n(S^n - f^{-1}(y))$ onto a subgroup of $\mathbb Z$ and passing to quotient, we have $$H_n(S^n, S^n - f^{-1}(y)) \approx \frac{\mathbb Z}{\xi(H_n(S^n - f^{-1}(y)))}.$$ Since all subgroups of $\mathbb Z$ are $n\mathbb Z$ for some $n \in \mathbb Z$, we have $H_n(S^n - f^{-1}(y))\approx \mathbb Z_n$, which gives a contradiction.
I can't find the mistake in this argument (that can be done for all $A\subset S^n$, and then all groups $H_n(S^n,A)$ are isomorphic to $\mathbb Z_n$ for some $n$), but I believe that it is just a detail that I didn't see.
Can anyone see the mistake?