I was wondering if anyone could give me some intuition as to why, for a smooth projective variety $X$ over $\mathbf{C}$ of complex dimension $d$, the Tate twist on $H^n(X(\mathbf{C}),\mathbf{Z})$ to be incorporated to have a pairing with $H_{d-n}(X(\mathbf{C}),\mathbf{Z})$ into $\mathbf{Z}(-d)$, is related to (tensor powers of) the cohomology group $H^2(\mathbf{P}^1,\mathbf{Z})\simeq H^1(\mathbf{G}_m,\mathbf{Z})$.
In other words, why does one often define $\mathbf{Z}(-1) := H^2(\mathbf{P}^1,\mathbf{Z})$? Is there some Kunneth formula lurking behind the scene? Should I think about $X(\mathbf{C})$ as a family $X(\mathbf{C})\times \mathbf{P}^1\to \mathbf{P}^1$? Is there a map $H^2((\mathbf{P}^1)^d,\mathbf{Z})\to H^d(X(\mathbf{C}),\mathbf{Z})$? Where does it come from?
I can understand the notation comes from $H_0(\mathbf{P}^1,\mathbf{Z})\simeq H_1(\mathbf{G}_m,\mathbf{Z})\simeq\pi_1^{\rm ab}(\mathbf{G}_m) =\pi_1(\mathbf{G}_m) := \mathbf{Z}(1)$, the reason I'm asking the question: even if there is $H_0(X(\mathbf{C}),\mathbf{Z})\to H_0(\mathbf{P}^1,\mathbf{Z})$, I see no map $X(\mathbf{C})\to\mathbf{P}^1$ in general. Unless somehow $X(\mathbf{C})$ can always be realized as the fiber over some point of $\mathbf{P}^1$ of some map $\mathcal{X}\to\mathbf{P}^1$, with $X$ homotopy equivalent to $\mathcal{X}$. Can one just choose $\mathcal{X} = X(\mathbf{C})\times\mathbf{P}^1$? (can it possibly be so simple? In algebraic geometry, this would amount to taking the trivial deformation of $X\to *$ to $\mathcal{X} \to\mathbf{P}^1$ along a fixed point $*\to\mathbf{P}^1$, and trivial deformations strike me as usually not so interesting). I'll appreciate any insight on the matter a lot
Thanks