When are Linear Operator and Identity transversal in a Vector Space?

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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

Since

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)

We have

$A: V \to V$ 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 ?

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Submersions are surjective on tangent spaces, by definition. This means the image of $df$ is the entire tangent space. In particular, no matter what other subspace you add to it, you will get the entire tangent space. This means it is transverse to everything.

I am not sure you are asking the question you intend to ask, however, since transverse intersection is a concept that applies to maps between manifolds, and yet you started out with just a linear operator on a vector space. Are you intentionally treating this vector space as both a manifold and identifying it with the tangent spaces? Or are you doing something else? If this is a homework question, I suggest you state this and post the question as close to verbatim as possible to avoid introducing notational mistakes.