Why is a transversal intersection of submanifolds a manifold?

Question

Let $N, M \subseteq R^n$be transversal submonifolds of $R^n$ We say that $N$ and $M$ are transversal if $T_pN + T_pM =T_pR^n$ for all $p \in N\cap M$. Why is $N \cap M$ a manifold?

Answer 1

Let $p$ be the codimension of $M$ and let $q$ be the codimension of $N$. Let us fixed $x\in M\cap N$, there exists $f=(f_1,\cdots,f_p)\colon\mathbb{R}^n\rightarrow\mathbb{R}^p$ and $g=(g_1,\cdots,g_q)\colon\mathbb{R}^n\rightarrow\mathbb{R}^q$ such that $\mathrm{rg}(\mathrm{d}_xf)=p$ and $\textrm{rg}(\mathrm{d}_xg)=q$. Moreover, $f=0$ is an equation for $M$ in a neighborhood of $x$ and $g=0$ is an equation for $N$ in a neighborhood of $x$. Let define $\varphi:=(f_1,\cdots,f_p,g_1,\cdots,g_q)\colon\mathbb{R}^n\rightarrow\mathbb{R}^{p+q}$, then $\varphi=0$ is an equation for $M\cap N$ in a neighborhood of $x$. Besides, the hypothesis $T_xM+T_xN=\mathbb{R}^n$ implies that: $$\dim(\ker(\mathrm{d}_xf)\cap\ker(\mathrm{d}_xg))=n-\dim\ker(\mathrm{d}_xf)-\dim\ker(\mathrm{d}_xg).$$ Which can be written as: $$\dim(\ker(\mathrm{d}_x\varphi))=n-p-q.$$ Therefore, $\varphi$ has rank $p+q$. Whence the result.