Let $f:N \to M$ be a smooth constant rank map between smooth manifolds.
The rank theorem says that such a map can be injective only if its rank is equal to the dimension of $N,$ which in turn has to be smaller or equal to that of $M.$
Say that some smooth map $f$ is an immersion when the rank of $f$ at a point $p$ is equal to the dimension of the domain for every point $p.$
From the above claim, we have that any injective smooth map of constant rank must be an immersion.
On the other hand, I find the definition of globally injective immersion as a smooth map that is an immersion which is also bijective onto its image.
But any constant rank injective smooth map must automatically be both bijective onto its image and also an immersion.
So isn't the definition of globally inctive immersion just saying that $f$ is a constant rank smooth injective map? If yes, what's the use of this definition?
Best Answer
Yes, you gave a correct interpretation of an injective immersion. What your definition is good for, I do not know: Personally, I do not find it particularly useful. However, with immersions, the point is that they do not have to be injective. Furthermore, it is often easier to check that a map is an immersion than to verify injectivity: Verifying that a map is an immersion boils down to understanding the derivative of a map and then injectivity of linear maps.
Just think of functions of one variable. Suppose that somebody asked you to show that a smooth function $f$ is injective, say, $$ f(x)=x^5+ 2x+10. $$ This sounds like a hard task. But you can use the 1st derivative test $$ f'(x)=5x^4+ 2 $$ and notice that $f'(x)> 0$ for all $x$, implying that $f$ is strictly monotonic (increasing). Hence, $f$ is also 1-1. In terms of immersion, what you prove is that $f$ is an immersion since $f'(x)\ne 0$ for all $x$. Hence, $f$ is strictly monotonic (assuming it is defined and is differentiable on an interval). This method of proving injectivity fails in higher dimensions, so one settles for immersions rather than injective maps.