Show that a function is surjective on a ball

Question

Let $B(0,1)\subseteq\mathbb R^n$ be the ball of radius 1 at the origin. Suppose $f:B(0,1)\to\mathbb R^n$ satisfies $f(0)=0$ and $$\forall x\neq y\in B(0,1), |f(x)-f(y)-(x-y)|<0.1|x-y|$$ How can I show that for every $z\in B(0,0.4)$, there exists $x\in B(0,1)$ such that $f(x)=z$?

If $z\in B(0,0.4)$, then $|f(z)-z|\leq 0.04$ but I'm not sure how to show surjectivity from the conditions.

Answer 1

Hint

As Op and @greg Martin pointed out that $f$ is continuous.

Apply this property of continuous map ( hopefully you would have proved it )

under a continuous map , inverse image of an open set is open.

Notice that $B(0,0.4)$ is open set So, what will be $f^-1(B(0,0.4))$ ?

Explanation

$f^{-1}(B(0,0.4))$ is an open subset of $B(0,1)$, let us denote it by $\beta$. Then given a point $z$ in $B(0,0.4)$ then $f^{-1}(z)=x\in \beta$. ( For some $x$) Thus, $f(x)=z$ and this is true for every $z\in B(0,0.4)$. And oviously $x$ will be in $B(0,1)$.

Hope you can take it from here.