Let $I^3 : [0, 1]^3 \to \mathbb{R}^3, I^3(x) = x$. Define $I^3 _{(i, \alpha)} : [0, 1]^2 \to [0, 1]^3, \alpha \in \{0, 1\}$ like that : $$I^3_{(1, a)}(x^1, x^2) = I^3(\alpha, x^1, x^2), I^3_{(2, a)}(x^1, x^2) = I^3(x^1, \alpha, x^2), I^3_{(3, a)}(x^1, x^2) = I^3( x^1, x^2, \alpha).$$
I would like to compute pullbacks $(I^3_{i, \alpha})^*(dx^1 \wedge dx^2)$, however I am lost in what is happening here. How could I start computing something simpler eg. $(I^3_{1, \alpha})^*(dx^2)$?
$(I^3_{(1,\alpha)})^*dx^1 = 0 \\ (I^3_{(1,\alpha)})^*dx^2 = dx^1 \\ (I^3_{(1,\alpha)})^*dx^3 = dx^2 \\ (I^3_{(2,\alpha)})^*dx^1 = dx^1 \\ (I^3_{(2,\alpha)})^*dx^2 = 0 \\ (I^3_{(2,\alpha)})^*dx^3 = dx^2 \\ (I^3_{(3,\alpha)})^*dx^1 = dx^1 \\ (I^3_{(3,\alpha)})^*dx^2 = dx^2 \\ (I^3_{(3,\alpha)})^*dx^3 = 0 \\ (I^3_{(1,\alpha)})^∗(dx^1∧dx^2)=0∧dx^1=0 \\ (I^3_{(2,\alpha)})^*(dx^1 \wedge dx^2) = dx^1 \wedge 0 = 0 \\ (I^3_{(3,\alpha)})^∗(dx^1∧dx^2)=dx^1∧dx^2 $