I'm a physics student and my understanding on differential geometry and manifolds is based on general relativity. Can you explain the concept of (smooth) submanifold intuitively? Is a smooth manifold embedded in higher dimensional Euclidean space a submanifold? If that manifold is curved then its geodesics are not geodesics of the Euclidean space but I think it would make sense to define that geodesics of a submanifold also have to be geodesics of the manifold. Does some definition include that?
Intuitively what are submanifolds
differential-geometrymanifoldssmooth-manifoldssubmanifold
Related Solutions
First of all, your definition 1 is a bit faulty, you should say that $rank(df)_x$ is constant equal to $m=dim(M)$ at every point $x\in M$. With this in mind, these are equivalent definitions. However, I do not like either one of these definitions and for several reasons. Spivak's definition - because it depends on a nontrivial theorem (the immersion theorem), while a definition this basic should not depend on anything nontrivial. Also, for the reason that you stated. More importantly, I do not like both definitions - because they utterly fail in other, closely related situations. For instance, if I were to define the notion of a topological submanifold in a topological manifold along these lines, Spivak's will fail immediately (what is the rank of the derivative if I do not have any derivatives to work with?); Carroll's definition will fail because it will yield in some cases rather unsavory objects, like Alexander's horned sphere in the 3-space. The same if I were to use triangulated manifolds and triangulated submanifolds, algebraic (sub)varieties and analytic (sub)varieties.
Here is the definition that I prefer. First of all, what are we looking for in an $n$-dimensional manifold $N$ (smooth or not): We want something which is locally isomorphic (in whatever sense of the word isomorphism) to an $n$-dimensional real vector space (no need for particular coordinates, but if you like, just $R^n$). Then an $m$-dimensional submanifold should be a subset which locally looks like an $m$-dimensional vector subspace in an $n$-dimensional vector space. This is our intuition of a submanifold in any category (smooth, topological, piecewise-linear, holomorphic, symplectic, etc) we work with. Once you accept this premise, the actual definition is almost immediate:
Definition. Let $N$ be a smooth $n$-dimensional manifold. A subset $M\subset N$ is called a smooth $m$-dimensional submanifold if for every $x\in M$ there exists an (open) neighborhood $U$ of $x$ in $N$ and a diffeomorphism $\phi: U\to V\subset R^n$ ($V$ is open) such that $\phi(M\cap U)= L\cap V$, where $L$ is an $m$-dimensional linear subspace in $R^n$. (If you like coordinates, assume that $L$ is given by the system of equations $y_1=...y_{n-m}=0$.)
This is completely intrinsic. Next, you prove a lemma which says that such $M$ has a natural structure of an $m$-dimensional smooth manifold with topology equal to the subspace topology and local coordinates near points $x\in M$ given by the restrictions $\phi|(U\cap M)$. Then you prove that with this structure, $M$ satisfies the other two definitions that you know.
Remark. Note that this definition will work almost verbatim if I were to deal with topological manifolds: I would just replace "a diffeomorphism" with "a homeomorphism. If I were to work with, say, complex (i.e. holomorphic) manifolds, I would replace $R^n$ with $C^n$ (of course), use complex vector subspaces and replace "diffeomorphism" with "a biholomorphic map". An so on.
Now, to the question why is it so much more complicated than the concept of a subgroup or a submodule or any other algebraic concept you can think of. This is because manifolds have a much richer structure. To begin with, they are topological spaces. (Notice that every submanifold is equipped with the subspace topology, so this has to be built in.) Then, the notion of vector spaces has to be used at some point. Next, there is the "local" thing (local charts)....
For (3):
Assuming you allready know that the rank of $f$ is maximal on $f(M)$:
The rank $$\text{rk}\, f:M\to\mathbb Z$$ is lower-semicontinous and hence locally only can jump up.
This follows from the lower-semicontinuity of the rank of $n\times n$-matrices. So the set
$$\{x\in M:\text{rk}_xf=r\}=\{x\in M:\text{rk}_xf>r-\frac 12\}$$ is open and hence an open neighborhood of $f(M)$.
For (4):
Let $x_0\in f(M)$. Then by the constant rank Theorem there are charts $(U,\phi)$, $(V,\psi)$ about $x_0=f(x_0)$ with $f(U)\subseteq V$ and
$$\psi\circ f\circ \phi^{-1}(x_1,\dots,x_n)=(x_1,\dots,x_r,0,\dots,0)$$
and by shrinking $V$ one can assume $\psi(f(U))=\psi(V)\cap(\mathbb R^r\times\{0\}$).
Now set $W=U\cap V$, which is an open neighborhood of $x_0$. $f\circ f=f$ implies $W\cap f(M)=W\cap f(U)$ and $\psi$ injective implies $\psi(W\cap f(U))=\psi(W)\cap\psi(f(U))$, so
$$\psi(W\cap f(M))= \psi(W\cap f(U))= \psi(W)\cap \psi(V)\cap (\mathbb R^r\times\{0\}))= \psi(W)\cap(\mathbb R^r\times\{0\})$$
which shows that $f(M)$ is a submanifold of $M$.
Best Answer
If $M$ is a smooth manifold, then a submanifold of $M$ is a subset $N \subset M$ that is itself a manifold with the smooth structure it inherits from $M$. By the Nash embedding theorem, any manifold is a submanifold of Euclidean space, so in some sense that's the only case we need to consider.
Everyone's favorite submanifold is the sphere $S^n = \{ x \in \mathbb{R}^{n+1} \mid |x| = 1\} \subset \mathbb{R}^{n+1}$. For other cheap examples, you can look at the zeros of polynomials on $\mathbb{R}^n$, which will "usually" define a smooth submanifold.
The sphere shows you that the geodesics of the submanifold (great circles) do not need to be geodesics of the ambient manifold (straight lines). This can happen, however, if the inclusion map $j : N \to M$ is a Riemannian embedding. That is a much stronger condition than for $N$ to be a submanifold of $M$. The second fundamental form measures the failure of this condition; if it is zero, then the geodesics of the submanifold are geodesics of the ambient manifold, otherwise they will differ in general.