Let $A:V \to V$ a Linear Operator on the Vector Space $V$ and $Id:V \to V$ the Identity map, when is $A$ transverse to $Id$ ? My intuition says that $A$ may not have $1$ as eigenvalue but I didn't find out if it is really true
Two maps $f: X \to Z$ and $g:Y\to Z$ are transversal if, for every $ x \in X$ and $y\in Y$ with $f(x) = z = g(y)$, the diferentials on these points spans the entire tangent space at $z$ in sense that $$im(df) + im(dg) \simeq T_zZ $$ (by https://ncatlab.org/nlab/show/transversal+maps)
$A: V \to V$ and $Id:V\to V$ are transversal if, for every $x \in V$ and $y\in V$ with $A(x) = z = Id(y)$ , the diferentials at these points spans the entire tangent space at $z$ in sense that $$im(A) + im(Id) \simeq T_zV \simeq V$$ , i.e.
$$A(x) + Id(y) = z \ \ \forall \ \ x, y \in V \ s.t. \ \ A(x) = z = Id(y) , \ i.e. \\ \iff (A - I)(x) \ = x \ \ * \ for \ x \ = y \ = z$$
Since $(A-\lambda I)$ is singular $\iff det(A-\lambda I)=0$ $\iff \lambda $ is eigenvalue of $A$ , if $1$ is eigenvalue of $A$ then equation $\ \ * \ \ $ doesn't means isomorphism and then, by if and only if, doesn't means transversality of $A$ and $Id$.
My first problem is:
I assumed at equation $\ \ * \ \ $ the equality $x \ = y\ \ =z$, which makes sense to me because $A$ and $Id$ are global maps and then the transversality must be with all points of $V$ but I'm feeling that something is missing. $z \ = y \ $ because $Id$, no doubt, but $ x \ = y \ $?
My second problem is:
At https://ncatlab.org/nlab/show/transversal+maps the author says
'In particular, a submersion is transversal to all functions.'
But isn't Identity and non-singular Linear Operator surjective , equals to they derivative , because they are their best linear approximation, and then are submersions ?