Suppose we are trying to compute the homology of the space $\mathcal{T}$ with integer and $\mathbb{Z}/2$ coefficients, where $$ \mathcal{T} = \{ (a, b) \mid a,b \in S^n, \langle a, b \rangle = 0 \}$$ and $S^n$ is the $n$-sphere and the standard inner product is used. The hint I was given was to express an element of $\mathcal{T}$ as $(a_0,... , a_n, b_0, ..., b_n)$ and define $\mathcal{T}_+ = \{ (a,b) \in T \mid a_0 \ge 0\}$, $\mathcal{T}_- = \{ (a,b) \in T \mid a_0 \le 0\}$ and $$S_+ = \{ v \in S^n \mid a_0 \ge 0\} \ \ \ S_- = \{ v \in S^n \mid a_0 \le 0\}.$$ From there, I should show that $\mathcal{T}_{\pm} \simeq S_{\pm} \times S^{n-1}$. However, what's troubling me is that I don't even believe (technically or intuitively) that this homeomorphism is true. How should I even picture the space $\mathcal{T}$? Should I think of $\mathcal{T}$ as two $n$-spheres that are orthogonal? So for $n = 1$, $\mathcal{T}$ is two orthogonal circles? Are $S_+$ and $S_-$ just "hemispheres" of $S^n$?
Ignoring this lack of intuition, say I just want to find a homeomorphism. It seems pretty clear that I can map $$(a,b) \in T_+ \longmapsto (a, ???)$$ since $a \in S^n$ satisfies the property that $a_0 \ge 0$ and thus $a \in S_+$. But how would I even map $b = (b_0, ..., b_n) \in S^n$ to an element of $S^{n-1}$ (i.e. fill in the '???' part)? I don't see how to do that without using some type of projection, which definitely won't be a homeomorphism. I realized that the intersection $S_+ \cap S_-$ will be homeomorphic to $S^{n-1}$ (right?), but is that useful?