I have the following problem (which involved the Inverse Function Theorem and Injectivity):
Let $f(x_1,x_2,x_3)=(u(x_1,x_2,x_3),v(x_1,x_2,x_3),w(x_1,x_2,x_3))$ be the mapping of $\mathbb{R}^3$ and $\mathbb{R}^3$ given by $u=x_1$, $v=x_1^2+x_2$, and $w=x_1+x_2^2+x_3^3$. The Jacobian of $f$ is given by $$ \begin{pmatrix} 1 & 0 & 0 \\ 2x_1 & 1 & 0 \\ 1 & 2x_2 & 3x_3^2 \end{pmatrix} $$ which has determinant $3x_3^2$. As all the partials here are continuous, we know that $f$ is $C'$ on all of $\mathbb{R}^3$. Moreover, $f$ satisfies all of the conditions of the Inverse Function Theorem except for when $x_3=0$.
It is clear that $f$ is injective. Suppose $f(a,b,c)=f(x,y,z)$. The first coordinate gives $a=x$, using this in the second gives $b=y$, then using this in the third gives $c=z$ so that $f$ is everywhere injective.
The problem itself was simple enough. However, I have noticed many problems involving the Inverse Function Theorem and injectivity. But I'm not seeing the exact connection. Satisfying the Inverse Function Theorem gives only that there is some neighborhood on which $f$ is injective and hence has an inverse (which happens to be continuously differentiable). Why then all the questions connecting injectivity and the Inverse Function Theorem? Could one actually show a function is injective using the Inverse Function Theorem? Or is one always forced to check 'by hand' using something similar to what I did above?