I do not how to approach the computation of the homology of the following space
$X = \{ (z_1, z_2, z_3) \in (S^1)^2 \times D_2 : z_1 + z_2 + z_3 = 0 \}$
$S^1$ denotes the unit circle and $D_2$ the unit ball in $\mathbb{R}^2$
From the regular value theorem, $X$ is a compact orientable connected manifold of dimension $2$, so $H_0(X) = H_2(X) = \mathbb{Z}$.
But what can I say about $H_1$ (or the number of $1$-dimensional holes)?
Correct me if I am wrong (my smooth manifold knowledge is not great), but won't the preimage have boundary, so you cannot use Poincare duality?
I mention this because I believe the manifold is a cylinder, so my calculation of the homology will disagree with you. I believe the following is a deformation retraction to a circle (so its 2 dimensional homology would be 0).
You can view X as a subspace of $T^2$ by forgetting the third coordinate since it is fully determined by the first two. Let's call it $X' \subset T^2$ and it is given by pairs $(x,y) \in T^2$ such that the magnitude of their sum is at most 1. Let $F: X' \times I \rightarrow X'$ be given by $(x,y,t) \rightarrow (\frac{(1-t)x - ty}{|(1-t)x-ty|},y)$. The denominator is never zero because $y \neq x$, and it isn't hard to convince yourself that it lies inside $X'$ (though I'm sure that there is an argument with inequalities).
Thus, there is a deformation retraction to the anti-diagonal of $T^2$ (pairs of coordinates such that the first is the negative of the second) which is homeomorphic to $S^1$, so $H_1(X)=\mathbb{Z}$.