Is $f:\mathbb{R} \to [0,+\infty): x \mapsto 1+x^2$ not injective? So it's not surjective because if we pick a value $0$ which is part of the co-domain then it turns out that there is no such value that $x$ can take since the smallest $x^2+1$ can take is $1$ if $x = 0$. 
An injective function is a one-to-one function so for every value that is put in, i'll get a different value out. If any real number is put in then that means on a graph the bottom $2$ quadrants are excluded. If what can come out is between $0$ and infinity then if we pick out a value like $1$ and $-1$ to put in we get out respectively $2$ and $2$ which makes this function a $2$ to $1$ function in which case this function is then not injective.
Am i right?
 A: Your are right.  Also, note that
$1 + (x_0)^2 = 1 + (-x_0)^2 \tag 1$
for any $x_0 \in \Bbb R$.
A: Should be a "little" comment but I try to make it very quick for you to see it: $f(-1) = 2 = f(1)$. Thats all you got to show....
A: $f$ is not surjective
$\exists y \in[0,+\infty), \forall x\in \mathbb{R}, \quad y\ne f(x)$
$y=0<f(x),\quad \forall x\in\mathbb{R}$

if $f$ is injective
$\forall x_1,x_2 \in \mathbb{R}_+,\quad f(x_1)=f(x_2)\implies x_1=x_2 $
$(x_1)^2+1=(x_2)^2+1 \implies (x_1)^2=(x_2)^2\implies x_1=x_2 \quad \text{or}\quad \boxed{x_1=-x_2}$  contradiction
so $f$ is not injective
A: As others have said $f:\mathbb{R}\rightarrow [0,\infty)$ $f(x)=1+x^2$ is not injective. However, suppose we defined $f:[0,\infty)\rightarrow [0,\infty)$. Now the function is injective. In general, we can restrict the domain sufficiently of any function and force it to be injective. How much of the domain to shrink depends on both the function and the range. 
In the trivial case, just define $f:\{x\}\rightarrow\{y\}$ where $x,y\in \mathbb{R}$, and any function would actually be continuous, injective, and even surjective. 
