Let $X = D^3 \cup_f S^2$ where $f: \delta D^3 \cong S^2 \rightarrow S^2$ is a degree d attaching map. Compute $H_i(X)$
$proof:$ I will use the mayer vietoris sequence.
Let $A = S^2$ and $B=D^3$. Then $A \cap B \cong S^2$ and our sequence will look like:
$0 \rightarrow H_2(S^2) \rightarrow^a H_2(D^3) \oplus H_2(S^2) \rightarrow^b H_2(X) \rightarrow^c H_1(S^2) \rightarrow 0$
Making the obvious substitutions we have:
$0 \rightarrow \mathbb{Z} \rightarrow^a \mathbb{Z} \rightarrow^b H_2(X) \rightarrow^c 0$
Where $a(1) = d$
Thus $H_2(X) \cong \frac{\mathbb{Z}}{d\mathbb{Z}} \cong \mathbb{Z}_d$
Is this correct?
This looks right to me. BTW, you can regard this as the "suspension" of the "same" sequence one dimension lower, where you have (in polar coordinates) $$ f: \partial D^2 \to S^1 : (1, \theta) \mapsto 2 \theta, $$ (for the case $d = 2$, which is the nicest one), which may be easier to understand. The resulting space $X$, in this simpler case, is just $\Bbb RP^2$; the nice thing is that the fundamental group (hence first homology) of $\Bbb RP^2$ is $\Bbb Z/ 2 \Bbb Z$. (More generally, for the degree-$d$ map, you get $\Bbb Z/ d \Bbb Z$, whose abelianization is still $\Bbb Z/ d \Bbb Z$.) [NB: you can compute the fundamental groups in these cases pretty easily via Seifert/van Kampen.]
The only remaining challenge is to figure out how this lower-dimensional-sequence can tell you anything about the one you're looking at. I suspect that the 5-lemma comes into play, with the vertical maps all being induced by suspension, but I haven't actually worked through the details.