Proving that sum of injective and Lipschitz continuous function is injective?

Question

Suppose that $H:A\rightarrow B$ can be written as $H(x) = x + h(x)$, where $h$ is Lipschitz on $A$ with constant $0 < \delta < 1$. I need to show that H is injective.

Here is my attempt at proving this please let me know if it is correct.

Proof: Suppose that H is not injective and hence $H(x) = H(y) \text{ for } x\neq y$. This can be written as $x+h(x) = y+h(y)$. Now consider a constant function such that $h(x) = h(y)$ which is obviously Lipshitz but this implies that $x=y$ which is a contradiction and hence $H$ is injective.

Answer 1

Suppose $H(x)=H(y)$, then $h(x)-h(y) = y-x$. Since $d(h(x),h(y)) \le \delta d(y,x)$, we get $d(x,y) \le \delta d(x,y)$ and the only solution to this is $x=y$.

Answer 2

I'd think of it this way: you have two real functions, one, say, $f$, is $(1 - \epsilon)$-contracting and other $g$ $(1 + \delta)$-expanding, $\epsilon > 0, \delta \ge 0 $ , and you want to prove injectivity of their sum. If you look at their graphs, then for each point on graph of $f$ will lie in a cone like this $><$ with apex on that point and graph of $g$ will lie inside vertical part of similar one, but with obtuse angle. With a bit of triangle inequality usage you'll see that $C(f + g)$ will be expanding for some $C$, expanding maps are obviously injective and multiplication is a bijection.