Why is an embedding an injective immersion? My course on manifolds defines an embedding as follows:
'A smooth map $f:\mathcal{M}\rightarrow\mathcal{N}$ between manifolds $\mathcal{M}$ of dimension $m$ and $\mathcal{N}$ of dimension $n$ is an embedding if it is a diffeomorphism onto its range. We refer to the range, $f(\mathcal{M})$, of such a map, as a submanifold of $\mathcal{N}$.'
Below the definition, it states 'clearly, any embedding is an injective immersion', but I am struggling to see why this is. Injectivity follows from the fact that any diffeomorphism is bijective, but why does the immersion part follow? Why does $f$ being a diffeomorphism onto its range imply that the derivative $d_xf$ is injective for all $x\in\mathcal{M}$? Is it to do with $d_xf$ having a left inverse?
I've tried to consider a dimension argument but this has got me even more confused, so I have a second question: we view the $d_xf$ as a linear map from $T_x\mathcal{M}$ to $T_{f(x)}\mathcal{N}$, which can be represented as an $n\times{m}$ matrix since the dimension of a tangent space is equal to the dimension of its corresponding manifold. But couldn't we equally view $d_xf$ as a linear map from $T_x\mathcal{M}$ to $T_{f(x)}f(\mathcal{M})$? This can then be represented as an $m\times{m}$ matrix, since the dimension of $f(\mathcal{M})$ is equal to the dimension of $\mathcal{M}$ (diffeomorphisms preserve dimension?). Which is the correct way to look at it?
 A: I'm going to take their definition and go backwards. Let's start with the definition of a manifold. It sounds like to you, a manifold means a subset $M$ of $\Bbb R^n$ such that for all $p \in M$, there is a diffeomorphism $\psi: \Bbb R^n \to U$, $U$ an open subset of $\Bbb R^n$ that contains $p$, such that $M \cap U = \psi(\Bbb R^k)$, where $\Bbb R^k$ is the subspace of $\Bbb R^n$ where only the first $k$ coordinates can be nonzero.
Now, let's define an embedding of manifolds. What they say is almost fine - provided they add the phrase "such that $f(M)$ is a manifold". That is, I think their definition should say "An embedding $f: M \to N$ is a smooth map such that $f(M)$ is a manifold and such that the map $f$ is, then a diffeomorphism onto its image." If we don't demand that $f(M)$ is a manifold, this just doesn't make sense. (For an example of where $f(M)$ is not a manifold, take eg the figure 8: I can make this the image of a smooth immersion from the circle. Also see, for instance, the graph of |x|, which I can make the image of an injective smooth map from $\Bbb R$ - just not an immersion.) 
Note, in addition, that this contains the demand that it be a topological embedding - a homeomorphism onto its image. An injective immersion is not good enough unless the map is also proper: take the figure 8 above, and then write it as the injective image of $\Bbb R$. (The two 'tails' of $\Bbb R$ approach the center point from the bottom left and top right.) Then this is obviously not a homeomorphism onto its image. But injective proper maps are automatically topological embeddings in this context. (One case where you don't want to consider proper maps: open subsets of $\Bbb R^n$, like $GL_n(\Bbb R) \subset \text{Mat}(n \times n)$!)
Given this, let's prove the claim. First, it had better be injective, because it's a topological embedding. Why is it an immersion? Let's go back to the definition of diffeomorphism: a diffeomorphism is, in particular, an immersion. So if it's a diffeomorphism onto its image, then the map $M \to f(M)$ is an immersion; and by the fact that $f(M)$ is a manifold sitting inside $N$, the map $f(M) \to N$ is an immersion. (If you want to be careful about the proof of this, think about the charts we were guaranteed in the first paragraph in the definition of manifold, and think about how they automatically imply that the inclusion map $f(M) \to \Bbb R^K$ is an immersion, where $\Bbb R^K$ is whatever Euclidean space $N$ sits inside.)
A: Maybe this explanation is a bit advanced, but the reason is really just the chain rule, i.e. the fact that the tangent bundle is functorial: $T(g\circ f) = T(g) \circ T(f)$, and $T(\operatorname{id}_M) = \operatorname{id}_{TM}$ for any smooth maps $f$ and $g$ which you can compose. Notice that by the chain rule $T$ takes diffeomorphisms to linear bundle isomorphisms, which of course restrict to linear isomorphisms of vector spaces if you fix a point on the base.
