I've been reading Benny Rushing's book Topological Embeddings. In the last section of the book he gives a topological proof (which he attributes to R. D. Edwards) that a punctured $n$-torus admits an immersion (i.e. a continuous map which is locally an embedding) in $\mathbb{R}^n$. Since my question is about a step in the proof, I've included a brief description of the proof below.
The proof is by induction on $n$, with the following inductive statement:
$n$-dimensional inductive statement: There exists an immersion $f$ of $T^n \times I$ into $\mathbb{R}^n \times I$ such that $f|T^n_0 \times I$ is a product map, $f=\alpha \times 1$, where $T^n_0$ is $T^n$ minus an $n$-cell.
The notation is as follows: $I=[-1, 1]=J$, $S^1=I \cup_{\partial} J$, $T^n=(S^1)^n$ and $T^n_0=T^n-\mathring{J^n}$. We regard $\mathbb{R}^n\times S^1$ as a subset of $\mathbb{R}^{n+1}$ where the $I$-fibers are straight and vertical in $\mathbb{R}^{n+1}$. Assume $f$ and $\alpha$ are given in dimension $n$, and extend $f$ to an immersion of $T^n \times I$ into $\mathbb{R}^n\times I$ by $f \times 1_{S^1}: T^n \times S^1 \times I \to \mathbb{R}^n \times S^1 \times I \subseteq \mathbb{R}^{n+1} \times I$. The map $f \times 1_{S^1}$ needs to be adjusted since it is only a product on $T^n_0 \times S^1 \times I \subseteq T^{n+1}_0 \times I$. Without loss of generality we may assume that $f(T^n \times [-\frac{1}{2}, \frac{1}{2}]) \subseteq \mathbb{R}^n \times [-\frac{2}{3}, \frac{2}{3}]$. Let $\lambda$ be a homeomorphism of $I^2$ that is the identity on $\partial I^2$ and is a $\frac{\pi}{2}$-rotation on $[-\frac{2}{3}, \frac{2}{3}] \times [-\frac{2}{3}, \frac{2}{3}]$. We can extend $\lambda$, via the identity, to a homeomorphism $\overline{\lambda}: S^1 \times I \to S^1 \times I$.
Consider now the following immersion $$h=(1_{\mathbb{R}^n} \times \overline{\lambda}^{-1})(f \times 1_{S^1})(1_{T^n} \times \overline{\lambda}): T^{n+1} \times I \to \mathbb{R}^{n+1} \times I.$$ If we let $g=h|T^{n+1} \times [-\frac{1}{2}, \frac{1}{2}]$, then $g$ is a product on $(T^n_0 \times S^1) \cup (J^n \times [-\frac{1}{2}, \frac{1}{2}])$, which is a deformation retract of $T^{n+1}_0 \times [-\frac{1}{2}, \frac{1}{2}]$. Thus, without loss of generality we can assume that $g$ is a product on $T^{n+1}_0 \times [-\frac{1}{2}, \frac{1}{2}]$.
Why can we assume the bold statement?