Proving bijectivity of $f:[0,\infty)\to[0,2)$, where $f(x)=\frac{1}{x}+1$ for $x>1$, and $f(x)=x$ for $x\le 1$ Let $f: [0, \infty) \rightarrow [0,2)$ be defined by $$f(x) = \left\{
\begin{array}{ll}
      \frac{1}{x} + 1, &amp; x &gt; 1 \\
      x, &amp; x\le 1 \\
\end{array} 
\right.$$
  
  The question asks to prove that the function is bijective.


I have to prove it is injective and surjective. 

Injectivity:
if $x,y&lt;1$, then $x=y$
if $x,y&gt;1$ then, $\frac{1}{x}+1=\frac{1}{y}+1$, then $x=y$


  I get this bit, but I don't get how to prove the surjective part of the proof. 


Please could you help?
Let $$f_1(x)=x$$ for $$x\in [0,1]$$
and
$$f_2(x)=\frac 1x +1$$ for $$x&gt;1$$

$f_1$ is a bijection from $[0,1] $ to $[0,1]$ and $ f_2$ is a bijection from $(1,+\infty)$ to $ (1,2)$. thus $f$ is a bijection from $[0,1]\cup (1,\infty)$ to $[0,1]\cup (1,2)$.

Other proof.

Let $ y\in[0,2)$. We will prove it has only one antecedent.

If $y\in[0,1]$ then $f(y)=y$.
the antecedent is unique in $[0,1]$

if $y\in(1,2)$ then $f(\frac{1}{y-1})=y$ with $\frac{1}{y-1}&gt;1$.
the antecedent is unique in $(1,\infty)$
Injectivity:  Suppose $f(x) = f(y)$.

There are four cases:  $0 \le x \le 1$ and $0\le y \le 1$ and $f(x) = x = f(y) = y$.  Thus $x = y$.

$0 \le x \le 1$ and $y &gt; 1$ and $f(x) = x = f(y) = \frac 1y +1$.  So $x = \frac 1y + 1$. And $y= \frac 1{x-1}$.  But if $y &gt; 1$ then $0 &lt; \frac 1y &lt; 1$ and $1 &lt; \frac 1y + 1&lt; 2$ which contradicts $0\le x \le 1$.  (Likewise $0\le x=\frac 1y + 1 \le 1\implies -1 \le \frac 1y \le 0\implies y &lt; 0$ also a contradiction).

$x&gt; 1$ and $0 \le y \le 1$.  .... also gives a contradiction...  Too similar to the above to bear discussing.

[Note: we could have avoided these contradictory options by noting that $0\le w \le 1 \implies 0\le w=f(w) \le 1$ and $w &gt; 1 \implies 0&lt;\frac 1w &lt; 1\implies 1 &lt; \frac 1w + 1 = f(w) &lt; 2$.  So $f(w) \in [0,1] \iff w \in [0,1]$ and $f(w) \in (1,2) \iff w\in (0,1)$.]

Fourth and final case $x &gt; 1$ and $y &gt; 1$.  Then $\frac 1x + 1 = \frac 1y + 1$ so $\frac 1x = \frac 1y$ so $x=y$.

So $f(x)=f(y)\implies x=y$. 

Surjectivity:

Let $k \in [0, 2)$.  If $1 &lt; k &lt; 2$ and if $2&gt;\frac 1x + 1 = k &gt;1$ then $1&gt; \frac 1x = k- 1 &gt; 0$ and $x =\frac 1{k-1} &gt; 1$.  

So if $1 &lt; k &lt; 2$ then $\frac 1{k-1} &gt;1$ so $f(\frac 1{k-1}) = k$.

If if $0\le k \le 1$ then $f(k) = k$.

Either way for all $k\in [0,\infty)$ there is $x$ either equal to $k$ if $k \le 1$ or equal to $\frac 1{k-1}$ if $k &gt; 1$  so that $f(x) = k$.