$f \in C^{\infty} (\mathbb{R}^3, \mathbb{R}^4)$, $f_1(x)^2+f_2(x)^2=1=f_3(x)^2 + f_4(x)^2$. Prove that pullback of $f$ on every 3-form is $0$.

We have $f \in C^{\infty} (\mathbb{R}^3, \mathbb{R}^4)$, $f = f_1, f_2, f_3, f_4$.

Elements of that transformation satisfy:

$$f_1(x)^2 + f_2(x)^2 = 1 = f_3(x)^2 + f_4(x)^2 \ \text{ for: } \ x \in \mathbb{R}$$

Prove that pullback using $f$ on every 3-form is equal to $0$.

$$f^*( \omega) = 0 \ \text{ for: } \ \omega \in \Omega^3(\mathbb{R}^4)$$

Solution:

We have $f: \mathbb{R^4} \to \mathbb{R}^4$.

I think that it can be said that both $f_1(x)^2 + f_2(x)^2 = 1$ and $f_1=3(x)^2 + f_4(x)^2 = 1$ are unit circles. Therefore we know that the product $f(\mathbb{R}^4)$ consists of $S_1 \times S_1 \subset \mathbb{R}^4$.

Because the image has dimension $2$ (because it consists of $2$ unit circles), the pullback as a 3-form on $\mathbb{R}^4$ is equal to $0$ because it can't be defined on image with dimension lesser than $3$.

My solution is based solely on intuition and is probably wrong. That's why I would like to ask for help.