Let $X$ and $Y$ be submanifolds of $\mathbb R^N$. Show that for almost every $a\in \mathbb R^N$, the translate $X+a$ intersects $Y$ transversally.
Here is my attempt to fill in the details in the sketch of proof outlined here: Confused with The Transversality Theorem when all manifolds are boundaryless.
The differential of the map $F: X\times \mathbb R^N\to \mathbb R^N$, $(x,a)\mapsto x+a$ is given by $DF_{(x,a)}(u,v)=u+v$ (because this map satisfied the definition of the differential and the differential is unique). Its kernel is $N$-dimensional because it consists of all elements of the form $(q,-q)$ with $q\in \mathbb R^N$, so by the dimension formula the differential is onto. Let $y=F(x,a)$. Then $Im(DF_{(x,a)})+T_y Y= T_y \mathbb R^n$ holds because $Im(DF_{(x,a)})=\mathbb R^n=T_y \mathbb R^n$, so any vector $v$ in $T_y \mathbb R^n=\mathbb R^n$ is the sum $v+0$ of the vector $v\in Im(DF_{(x,a)})=\mathbb R^n$ and $0\in T_y Y$. Hence $F$ is transverse to $Y$. By the theorem in the cited question, for almost every $a\in \mathbb R^n$, $f_a: x\mapsto F(x,a)$ is transversal to $Y$.
But we need to show that $T_x(X+a)+T_x(Y)=T_x \mathbb R^N$ for all $x\in (X+a)\cap Y$.
My questions are whether what I wrote above correct and how to conclude the proof.