If $M_1, M_2$ are two manifolds such that $M_1 \subset M_2$, show that $ \dim(M_1) \leq \dim(M_2)$

Question

We are given two manifolds $M_1 \subset M_2 \subset \mathbb{R}^n$. I need to prove that $\dim(M_1) \leq \dim(M_2)$.
This is part of calculus 4 course, and I haven't taken a course in topology yet, so the proof should be elementary.
I was able to show that if $k_1 = \dim(M_1), k_2 = \dim(M_2)$ then there exists an open subset $U \subset \mathbb{R}^{k_1}$, and another subset (which I don't think is necessarily open) $V \subset \mathbb{R}^{k_2}$, such that there exists an homeomorphism $T: U \to V$. Does that mean that $k_1 \leq k_2$ somehow?

Answer 1

Let $\dim{M_1}=p$, $\dim{M_2}=q$. Suppose we have smooth atlases $ \{ (U_\alpha ,\phi_\alpha) \}_{\alpha \in \mathcal{A}}$ and $ \{ (V_\beta, \phi_\beta \}_{\beta \in \mathcal{B}}$ mapping open sets in $M_1$ to open sets in $\mathbb{R}^p$, and those in $M_2$ to $\mathbb{R}^q$. Then, since $M_1$ is a submanifold of $M_2$ (since it is a subset with the structure of a manifold) we have an (injective) embedding given by an inclusion map $f:M_1 \rightarrow M_2$ so that, for any $x \in M_1$ $df_x: T_xM_1 \rightarrow T_{f(x)}M_2 = T_xM_2$ is injective. Thus note the linear map $d(\phi_\beta \circ f \circ \phi_\alpha^{-1}):\mathbb{R}^p \rightarrow \mathbb{R}^q$ is injective when restricted to appropriate open sets (since it is a compositions of injections).

The rank-nullity theorem then gives that $p = \dim{\mathbb{R}^p} = \dim \left(\ker{d(\phi_\beta \circ f \circ \phi_\alpha^{-1}}\right) +$rank$( {d(\phi_\beta \circ f \circ \phi_\alpha^{-1}}))$. But we know that the map is injective, so $ \dim \left(\ker{d(\phi_\beta \circ f \circ \phi_\alpha^{-1}}\right) = 0$, i.e. $p = $ rank ($ {d(\phi_\beta \circ f \circ \phi_\alpha^{-1}}))$. Finally, we note that the rank of a linear map is the dimension of its range, which in this case is a subset of $\mathbb{R}^q$. It follows that

$\dim M_1 = p = \dim{\mathbb{R}^p}$ = rank ($ {d(\phi_\beta \circ f \circ \phi_\alpha^{-1}})) \leq \dim{\mathbb{R}^q} = q = \dim M_2$.