What is the reason for this step in proving the Borsuk-Ulam Theorem (by triangulation)

Question

I'm reading Using the Borsuk-Ulam Theorem, which presents several proofs of the theorem. Since this is a book for combinatorialists, the first involves a triangulation $\mathsf{T}$ of the $n$-dimensional crosspolytope $C^n = \partial(\text{conv}\{\pm e_1,\dots,\pm e_n\})$, since proving it for $C^n$ is enough for $S^n$, as $C^n \cong S^n$.

Here is the proof sketch; I understand all the technical details, but I'm having trouble understanding why a to-be-specified step is taken.

1. Suppose for contradiction that an antipodal continuous $f : C^n \rightarrow \mathbb{R}^n$ has no zero. Let $\pi$ be the projection $\mathbb{R}^{n+1} \rightarrow \mathbb{R}^n$ which eliminates the last coordinate.

2. Consider the function $F(x,t) = (1-t)\pi(x) + tf(x)$ on $C^n \times I$, and notice that $F$ has two antipodal zeros on $C^n \times \{0\}$, the "north and south poles" of the crosspolytope.

3. Construct a very fine triangulation $\mathsf{T}$ of $C^n \times I$, and define $G$ to be $F$ on $V(\mathsf{T})$, and extend it to all of $C^n \times I$ affinely. (That is, if $x$ is a convex combination of vertices $\{v_i\}$, then $f(x)$ is that convex combination of the $\{f(v_i)\}$). Moreover, require that $\mathsf{T}$ is antipodal; each simplex $\sigma \in \mathsf{T}$ bijects to an antipodal simplex $a(\sigma) \in \mathsf{T}$. Since $\mathsf{T}$ is fine enough and $f$ has no zeros on the top sphere $C^n \times \{1\}$, we know that $G$ has no zeros on the top sphere.

4. Let $H$ be a antipodally perturbed $G$, and with small enough perturbations, the zero set of $G$ is in the zero set of $H$.

5. Notice that on each simplex $\sigma$, the zero set of $G$ must either be empty, or a line segment from one facet to another, not intersecting a lower-than-facet-dimensional face.

6. Notice that $H$ has the same property; that is, the bad perturbations which introduce zeroes on low-dimensional faces are a null set.

7. Conclude that the zero set of $H$ is locally polygonal, comprised of cycles, and one polygonal path going from the north pole on $C^{n} \times \{0\}$ to the south pole on $C^n \times \{0\}$. This is impossible, as the path is symmetric, so its midpoint must be a fixed point in the inversion map $(x,t) \mapsto (x,t)$, but such a map has no fixed point in this construction. Thus $f$ has a zero.

Overall, the proof method is "proof by linear approximation," where a complicated function defined on a shape is well approximated by a linear approximation on a triangulation of that shape.

I'm confused about steps (4) and (6). Why do we introduce the perturbation in the first place? I suspect that it's to guarantee that $f$ can be arbitrarily well-approximated by $H$, but I'm not really sure-- wouldn't a fine enough triangulation suffice?