Doubt in proof that retract of connected subsets of manifolds are manifolds

Question

I just saw a proof of this theorem :

Let $M$ be a connected smooth manifold and suppose that $f:M\rightarrow M$ is smooth with $f\circ f=f$. Then the image of $M$ will be a submanifold. That can be found in section $1.15$ here https://www.mat.univie.ac.at/~michor/dgbook.pdf.

Now I have 2 doubts about the proof :

1. He writes that for $x\in f(M)$ we have that $T_xf\circ T_x f= T_x f$. Now for this to be true wouldn't we need to say that there is an open set around $x$ such that all the elements in there are in $F(M)$? It seems that if we just have that one point that wouldn't work, since the derivative seems to be something local .

2. By linear algebra we have the rank-nulity theorem that gives us that $ rank (T_x f)+ \dim (\ker (T_x f))= \dim M$ . I understand the construction that he does and it based on the rank-nulity theorem. So we know the rank of a function does not need to locally constant , then in the proof the author uses the fact that $rank T_xf + rank(Id-T_xf)=dim M$ to prove that $rank T_xf$ will be locally constant. How does this happen ?

Thanks in advance.

Answer 1

Im not sure that I understand your first question.

Regarding your second question. Think about $f:\mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^2$. The rank is locally constant downwards if $rank(df_p)=1$ then locally around $p$ the rank is $1$. At the origin $p=0$, it is not the case. There is no open neighborhood $U$ around $0$ with $rank(df_x)=0$ for all $x \in U$. This is an example that the rank is not locally constant.

To understand where your reasoning fails think about $v_1,v_2,v_3 \in \mathbb{R}^3$ where $v_1=v_2$ and $v_1,v_3$ are linearly independent. the matrix $$A=[v_1,v_2,v_3]\in M_3(\mathbb{R})$$ is singular and $rank(A)=2$. Moreover if you change the first coordinate of $v_2$ by $\varepsilon$ small enough the matrix $$A_\varepsilon =\begin{bmatrix}v_1, & v_2+\varepsilon\cdot e_1 &,v_3\end{bmatrix}$$ will be nonsingular. Hence $\det(A_\varepsilon)\ne 0$. Its true that the determinant is continuous but $\{0\}$ is a closed set hence a singular square submatrix in the Jacobian can turn non singular.