Showing that the differential is an immersion

Question

If $f: X \rightarrow Y$ is an immersion of smooth manifolds, then show that $df: TX \rightarrow TY$ is also an immersion.

The definition of immersion(when dim$X <$ dim$Y$) that I have is that for $f: X \rightarrow Y$, $f$ is an immersion if $df_{x}: TX \rightarrow TY$ is injective. So, for my problem I suppose I would have to show that the differential of $df$ is also injective. But, how do I go about showing that?

Answer 1

Here's a hint: Note that $f:X \to Y$ is an immersion if and only if for any $x \in X$, you can find chart maps $(U,\alpha)$ and $(V, \beta)$ such that $x \in U$ and $f(x) \in V$ and such that $\beta \circ f \circ \alpha^{-1} = \iota$ is the inclusion mapping from $\Bbb{R}^{\dim X} \to \Bbb{R}^{\dim Y}$. The "if" part is trivial, and the "only if" can be deduced as a consequence of the inverse/implicit function theorem.

Now, $d\beta \circ df \circ (d\alpha)^{-1} = d \iota$, so just try to verify that $d \iota$ can be regarded as an inclusion mapping between (open subsets of) $\Bbb{R}^{2 \dim X} \to \Bbb{R}^{2 \dim Y}$. Once you do this, by the above characterization of immersions, you'll be done.

---

Basically the idea is that the theorem is obviously true for the inclusion mapping $\iota$ of vector spaces. But locally (by the implicit function theorem), every immersion "looks like" such an immersion. A similar idea works with submersions.

Answer 2

In our case, $df_{x}$ is a linear map from the vector space $T_{x} X$ to $T_{f(x)} Y$. Recall that the derivative of a linear map is itself (one way to think of the derivative of a function $f$, is that $df_{x}$ is the linear map which best approximates $f$ near $x$)

So, if $df_{x}$ is injective for $x \in X$, its derivative will be as well.