Prove that a ball with radius $r<1$ is a subset of the range of approximate identity function from a unit ball I am struggling with finding a proof (or counterexample) for the following Theorem:
Let $x\in\mathbb{R}^n$ and $f:\mathbb{R}^n\to \mathbb{R}^n$ a continuous function. Assume $\forall x\in B_1$, $||x-f(x)||_{\ell^2}<\varepsilon$ for some given $0\leq\varepsilon<1$. Then $B_{1-\varepsilon}\subset f(B_1)$, where $B_r:=\{x\in\mathbb{R}^n~|~||x||_{\ell^2}\leq r\}$ is the closed ball centered at $0$ with radius $r$.
Intuitively, it is clear that the unit ball can only compress up to an $\varepsilon$ when the function $f$ is applied. However, it is not directly clear that there are no 'holes' in the image $f(B_1)$. Continuity of $f$, together with local similarity intuitively takes care of this, but I don't know how to make it exact. 
I think the assumptions should be enough for giving the proof, but on top of them, I can say that $f$ is piecewise differentiable, $f$ is Lipschitz. There is no known point $x$ for which $||x-f(x)||=0$ AND $f(x)$ is differentiable. Also, the function $f$ is not injective. I believe it is surjective, but I'm not completely sure. 
 A: I think I found the answer myself, but I'm not 100% confident. Is there something I missed?
Let us first define a scaled function $\tilde{f}:\mathbb{R}^n\to \mathbb{R}^n$, with $\tilde{f}(x):=\frac{1}{1+\varepsilon}f(x)$. The range of $\tilde{f}$ is a subset of the unit ball $B_1$. 
The closed unit ball $B_1$ is a contractible space. By definition of a contractible space [Gamelin, 1999]: $\exists G: B_1 \times [0,1] \to B_1$ continuous such that $G(x,0) = x$ $\forall x$, and $G(x,1) = x_0$ $\forall x$. It can be seen that $\tilde{f}(B_1)$ is also a contractible space: $G(\tilde{f}(x),0) = \tilde{f}(x)$ $\forall x$, and $G(\tilde{f}(x),1) = x_0$ $\forall x$. This means that the scaled image under the unit ball can not have any 'holes' in it. Since $\tilde{f}$ is just a scaled version of $f$, the same holds for the $f$. 
Left to show is that the boundary of $f(B_1)$ lies outside $B_{1-\varepsilon}$, so that we know that the whole ball $B_{1-\varepsilon}$ is in de range of $B_1$. For this we make use of [Rudin, 1976, Theorem 4.22]: since $f$ is a continuous mapping of a metric space $(\mathbb{R}^n,||~||_\ell^2)$ into a metric space $(\mathbb{R}^n,||~||_\ell^2)$, and the boundary of the unit ball ($\partial B_1$) is a connected subset of $\mathbb{R}^n$, Moreover, $\forall x\in \partial B_1$, $f(x)\in B_{1+\varepsilon}\backslash B_{1-\varepsilon}$. This implies that the boundary of the unit ball lies completely outside $B_{1-\varepsilon}$, which completes the proof.
A: Let $\Phi_y(x) = x - f(x) + y$. Then for$\Vert x\Vert\leq 1$ and $\Vert y\Vert \leq 1- \varepsilon$ we have,
$$
\Vert \Phi_y(x) \Vert \leq \Vert x- f(x) \Vert + \Vert y \Vert \leq \varepsilon + 1 - \varepsilon = 1
$$
Thus $\Phi_y$ maps $B_1$ to $B_1$ and is continuous, so by Brouwer fixed point theorem there exists $z\in B_1$ such that $\Phi_y(z) = z$. This implies for each $y\in B_{1-\varepsilon}$, there exists $z\in B_1$ such that
$$
z-f(z) + y = z \implies y = f(z)
$$
and thus we have shown that $y\in B_{1-\varepsilon}$ implies $y\in f(B_1)$
