Nontrivial cycles in the zero set of a map Let $B_4$ be a 4-disc, $X:=S^1\times B_4$, $\partial X=S^1\times S^3$ and $f: X\to\mathbb{R}^3$ be continuous, such that $f|_{\partial X}: (s_1, s_2)\mapsto H(s_2)$, where $H: S^3\to S^2\subseteq\mathbb{R}^3$ is the Hopf fibration. How to show that $H_1(f^{-1}(0))\stackrel{i_*}{\longrightarrow} H_1(X)$ is surjective (and is it necessarily true)? 
(I appologize that it looks so technical but it is closely connected to a deeper problem: if I know a sphere-valued map $f$ on the boundary of some space $X$, then the zero set of possible $\mathbb{R}^n$-valued extensions correspond to solutions of $1$-perturbations (wrt. some norm) of the equation $f(x)=0$. My goal is to describe the topology of the solution set(s). In this example, does it need to contain a nontrivial circle?)
 A: I think the result is not true.
First, note that since $B^4$ is perfectly normal, for some $\epsilon < 1/2$, we can find a function $g \colon B^4 \rightarrow [0,\epsilon/2]$ which is $0$ along the boundary $S^3$, $0$ at the two points $\left(\pm \frac{1}{2},0,0,0\right)$, $\epsilon/2$ on a small ball of radius $\epsilon$ around the origin, and otherwise not $0$ or $\epsilon/2$ (by an extended version of Urysohn's Lemma).
Now let $R_{\pi} \colon B^4 \rightarrow B^4$ be the rotation by $\pi$ around the first two coordinates. Then the function $h \colon B^4 \rightarrow [0,1]$ given by $\frac{1}{2}(g + g\circ R_{\pi}$ satisfies the same properties of $g$, but is also $R_{\pi}$-invariant.
Consider the function $\widetilde{f} \colon \mathbb{R} \times B^4 \rightarrow \mathbb{R}^3$ given by sending $(\theta,v=(v_1,v_2,v_3,v_4))$ to $$\frac{\|v+(1/2\cos \theta,1/2\sin\theta,0,0)\| \cdot \|v - (1/2\cos\theta,1/2\sin\theta,0,0)\|}{\sqrt{\frac{5}{4}-v_1\cos\theta-v_2\sin\theta}\cdot \sqrt{\frac{5}{4}+v_1\cos\theta+v_2\sin\theta}}H\left(\frac{f \cdot (0,0,0,1) + v}{\|f \cdot (0,0,0,1) + v\|}\right).$$ It is easy to check that this is well-defined (i.e. the denominators are never zero and the input to $H$ has norm $1$). Further, when $\|v\| = 1$, it is easy to check that this map is just the Hopf map $H$. And finally, this is invariant under $\theta \mapsto \theta + \pi$, so this descends to a map $$f \colon X \rightarrow \mathbb{R}^3$$ satisfying the necessary conditions. However, $f^{-1}(0)$ consists of points of the form $$(\theta, \pm(1/2\cos\theta,1/2\sin\theta,0,0))$$ as $\theta$ varies. In $X$, this is a circle which wraps around the $S^1$-coordinate twice, and in particular, we have $i_* \colon H_1(f^{-1}(0)) \rightarrow H_1(X)$ is a doubling map $\mathbb{Z} \rightarrow \mathbb{Z}$, and is therefore not surjective.
