Formal immersion gives the tangent normal-bundle identity

If $M=\mathrm{\varnothing }$, all total spaces and maps in the claimed sequence and splitting are empty, so exactness and the isomorphisms hold vacuously, with the stated normal-rank convention. Hence assume $M$ is nonempty, so $m\le n$, and fix $x_0\in M$. In a chart $U$ of $M$ around $x_0$, a trivialization of $TM{\mathrm{\mid }}_U$ and a trivialization of $TN$ over a chart containing $f(U)$, the section of [F1] is given by a smooth matrix function $A$ of rank $m$ at every point. Reordering coordinates we may suppose an $m\times m$ block $B(x)$ of $A(x)$ is invertible near $x_0$; then the image of $A(x)$ equals the image of the block matrix $\left(\genfrac{}{}{0px}{}{I_m}{C(x)}\right)$ with $C=A_2{B}^{-1}$ smooth, so the image is a smooth subbundle over $U$ with the columns of $\left(\genfrac{}{}{0px}{}{I}{C}\right)$ as a smooth frame.

The local frames of step 1.1 agree on overlaps because they span the same subspace at every point, so they glue to a smooth subbundle $F(TM)\subseteq f^{\ast }TN$ of rank $m$. The quotient ${\nu }_F:=f^{\ast }TN\mathrm{/}F(TM)$ is a smooth vector bundle by [L1], and the quotient map is a smooth bundle map over ${\mathrm{id}}_M$; composing the fibrewise isomorphisms $F_x:T_{x}M\to F(TM{)}_x$ with the inclusion gives a smooth bundle map $TM\to f^{\ast }TN$ with image $F(TM)$ and kernel $0$, so $0\to TM\stackrel{F}{\to }{f}^{\ast }TN\to {\nu }_F\to 0$ is a short exact sequence of smooth vector bundles; ranks give $\mathrm{rank}{\nu }_F=n-m$.

Let $C\subseteq f^{\ast }TN$ be a smooth complement of $F(TM)$, which exists by [L2]. Fibrewise, $F_x\oplus \mathrm{id}:T_{x}M\oplus C_x\to f^{\ast }T{N}_x$ is injective between spaces of dimension $n$, hence an isomorphism; it is smooth as a bundle map over ${\mathrm{id}}_M$, so it is a smooth bundle isomorphism $TM\oplus C\to f^{\ast }TN$ restricting to $F$ on $TM$. The quotient map restricts to an isomorphism $C\to {\nu }_F$ because $C\oplus F(TM)=f^{\ast }TN$ and $C\cap F(TM)=0$; composing the inverse of this isomorphism with the isomorphism above gives $TM\oplus {\nu }_F\cong f^{\ast }TN$, $(v,q(c))\mapsto F(v)+c$. The construction depends on the choice of $C$; the formula displays that dependence, and no complement is distinguished without further data, so the splitting is not canonical.

Choose a smooth bundle metric on $f^{\ast }TN$, which exists by [L2]. Its orthogonal complement $F(TM{)}^{\perp }$ is a smooth subbundle, is a complement of $F(TM)$, and is canonically determined by the metric; step 3.1 applied to it gives the orthogonal splitting, and applying step 3.1 to two different complements $C,C^{\mathrm{\prime }}$ exhibits both ${\nu }_F$-decompositions as isomorphic, since both are identified with the quotient.

For a genuine immersion $(f,F)=(f,df)$ the image is $df(TM)$ and the same sequence exhibits ${\nu }_{df}=f^{\ast }TN\mathrm{/}df(TM)$; when $f$ is an embedding this is the normal bundle of the embedded image by [L3], where the metric identification with the orthogonal normal bundle is precisely the construction of step 4.1.