Mistake using Lefschetz Duality I'm not sure where is the mistake in the following argumentation:
$$\mathbb{Z}\cong H^2(S^2)\cong H^2(D^2\times S^2)\cong H_2(D^2\times S^2,S^1 \times S^2)= H_2(S^2\times S^2) \cong \mathbb{Z}\oplus \mathbb{Z}$$
Second equality is homotopy invariance, third one is Lefschetz Duality, forth one is that relative homology corresponds to the quotient for nice spaces(actually we need to do it for reduced homology but everything is connected and dimension is positive).
Not sure what is going wrong here.
 A: Your reasoning is mostly correct, except for this single step:
$$H_2(D^2\times S^2,S^1 \times S^2)= H_2(S^2\times S^2)$$
You are right that $H_n(X,A)\cong\widetilde{H}_n(X/A)$ under a mild condition on $X,A$, e.g. $A$ is closed and has an open neighbourhood that deformation retracts onto $A$ (a.k.a. the good pair), which is satisfied here (via collar neighbourhood).
However $(D^2\times S^2)/(S^1\times S^2)$ is not $S^2\times S^2$, even up to homotopy. And you've pretty much provided a proof for that.
I'm not exactly sure what this quotient is. @MichaelAlbanese believes the quotient to be homotopy equivalent to $S^4\vee S^2$, based on Thom space construction which I didn't verify.
From comments:

$D^2/S^1=S^2$, so...

Yeah, the moral of this story is that collapsing boundary to a point does not play well with products.
A: As pointed out by freakish, the quotient $(D^2\times S^2)/(S^1\times S^2)$ is not homotopy equivalent to $S^2\times S^2$. Instead, I claim the quotient is homotopy equivalent to $S^4\vee S^2$.
If $E \to B$ is a vector bundle, its Thom space $\operatorname{Th}(E)$ is defined to be the quotient $\operatorname{Th}(E) := D(E)/S(E)$ where $D(E)$ and $S(E)$ are the disc bundle and sphere bundle of $E$ respectively. The trivial bundle $\varepsilon^n$ has Thom space $\operatorname{Th}(\varepsilon^n) = (B\times D^n)/(B\times S^{n-1})$. When $B = S^2$ and $n = 2$, this is precisely the quotient relevant to your computation.
Note that $D^n/S^{n-1}$ is homeomorphic to $S^n$ and has a natural basepoint $\ast = S^{n-1}/S^{n-1}$. There is a corresponding continuous surjective map $\phi : B\times D^n \to B\times S^n$ and the preimage of $B\times\{\ast\}$ is $B\times S^{n-1}$, so $\phi$ descends to a map $(B\times D^n)/(B\times S^{n-1}) \to (B\times S^n)/(B\times\{\ast\})$ which is a homeomorphism. Now, if $B_+$ denotes $B$ with a disjoint basepoint $\bullet$, i.e. $B_+ = B\sqcup\{\bullet\}$, and $\Sigma$ denotes reduced suspension, then we have
$$\Sigma^nB_+ = \dfrac{B_+\times S^n}{B_+\vee S^n} = \dfrac{B\times S^n\sqcup\{\bullet\}\times S^n}{B\times\{\ast\}\sqcup \{\bullet\}\times S^n} = \dfrac{B\times S^n}{B\times\{\ast\}}.$$
That is, the Thom space of the trivial rank $n$ bundle $\varepsilon^n \to B$ is homotopy equivalent to $\Sigma^nB_+$.
When $B = S^m$, we can see that the suspension $\Sigma(S^m)_+ = \Sigma(S^m\sqcup\{\bullet\})$ is homeomorphic to $S^{m+1}$ with two points identified. This is illustrated in the diagram below for $m = 1$.

This space is well-known to be homotopy equivalent to $S^2\vee S^1$, see Example $0.8$ of Hatcher's Algebraic Topology. The same argument provided by Hatcher shows that $\Sigma(S^m)_+$ is homotopy equivalent to $S^{m+1}\vee S^1$. As reduced suspension distributes over wedge products, we see that
$$\Sigma^n(S^m)_+ = \Sigma^{n-1}(\Sigma(S^m)_+) = \Sigma^{n-1}(S^{m+1}\vee S^1) = (\Sigma^{n-1}S^m)\vee(\Sigma^{n-1}S^1) = S^{m+n}\vee S^n.$$
That is, the Thom space of $\varepsilon^n \to S^m$ is homotopy equivalent to $S^{m+n}\vee S^n$.
In particular, the quotient $(D^2\times S^2)/(S^1\times S^2)$, which is the Thom space of $\varepsilon^2 \to S^2$, is homotopy equivalent to $S^4\vee S^2$. Therefore
$$H_2(D^2\times S^2, S^1\times S^2) \cong H_2\left(\dfrac{D^2\times S^2}{S^1\times S^2}\right) \cong H_2(S^4\vee S^2) \cong H_2(S^4)\oplus H_2(S^2) \cong \mathbb{Z},$$
as expected.
