A bundle map with rank drop is not a formal immersion

In the standard trivializations $T{\mathbb{R}}^2\cong {\mathbb{R}}^2\times {\mathbb{R}}^2$ the map $F$ reads $(x,\xi )\mapsto (x,\mathrm{diag}(1,0)\xi )=(x,({\xi }_1,0))$, a smooth map covering the identity whose restriction to each fibre is linear; by [F2] it is a smooth bundle map over $f={\mathrm{id}}_{{\mathbb{R}}^2}$.

At every $x\in {\mathbb{R}}^2$ the fibre map is $F_x=\mathrm{diag}(1,0):{\mathbb{R}}^2\to {\mathbb{R}}^2$, whose kernel contains the nonzero vector $(0,1)$; hence $F_x$ is not injective at any point.

By [F1] the pair $(\mathrm{id},F)$ therefore fails the defining fibrewise-injectivity condition at every point and is not a formal immersion, although the base map is even a diffeomorphism. Injectivity on a dense open set or off a proper closed subset gives no conclusion at the remaining points: if any such fibre is noninjective, [F1] excludes a formal immersion; if all fibres are injective, the condition is satisfied. Neither surjectivity of the base map nor linearity of $F$ replaces that condition.