mutually transverse embedded submanifolds, natural bundle surjections, direct sum, isomorphism Let $N$ be a manifold and let $M_1, \dots, M_n \hookrightarrow N$ be mutually transverse embedded submanifolds, so $M = \cap M_i$ is an embedded submanifold of $N$ with $\text{T}_m(M) = \cap T_m(M_i)$ inside of $T_m(N)$. Let $j_i: M \hookrightarrow M_i$ be the embedding. Using the natural bundle surjections $\text{N}_{M/N} \to j_i^*(\text{N}_{M_i/N})$, show that the resulting map$$\text{N}_{M/N} \to \bigoplus_i j_i^*(\text{N}_{M_i/N})$$to the direct sum is an isomorphism. (What does it say on fibers?)
In the special case $N = \mathbb{R}^n$ with its standard inner product, use the identification of normal bundles with orthogonal subbundles of tangent bundles to identify this isomorphism as an equality of subbundles of the pullback of $TN$ to $M$. (What does it say on fibers?) In words and pictures, explain the geometric meaning of this equality in terms of "directions perpendicular to submanifolds."
What are some hints to get started in the right direction? Any help would be appreciated, thanks.
 A: By Lemma 3.10 from these notes on universal and normal bundles, MATH 396. Universal Bundles and Normal Bundles, applied to the inclusions $M \to M_i \to N$ we have a bundle surjection $\text{N}_{M/N} \twoheadrightarrow j_i^*(\text{N}_{M_i/N})$ that on $m$-fibers is the quotient mapping $$\text{T}_m(N)/\text{T}_m(M) \twoheadrightarrow \text{T}_m(N)/\text{T}_m(M_i).$$ Passing to the bundle mapping to the direct sum, on fibers we get the natural map$$\text{T}_m(N)/\text{T}_m(M) \to \bigoplus_i \text{T}_m(N)/\text{T}_m(M_i).$$Since the $\text{T}_m(M_i)$'s are mutually transverse inside of $\text{T}_m(M_i)$'s are mutually transverse inside of $\text{T}_m(N)$ with intersection $\text{T}_m(M)$, the source and target have the same dimension and the map is injective, so it is an isomorphism.
In the special case $N = \mathbb{R}^n$ we get the orthogonal bundle $(TM)^\perp$ inside of $j^*(TN)$ and orthogonal bundles $(TM_i)^\perp$ inside of $k_i^*(TN)$ for each $i$ $($with $k_i: M_i \hookrightarrow N$ the embedding$)$. As such, $j_i^*((TM_i)^\perp)$ is contained in $(TM)^\perp$. Indeed, this can be checked on fibers, where it says that the subspace $\text{T}_m(M_i)^\perp$ in the inner product space $\text{T}_m(N) = \text{T}_m(\mathbb{R}^n)$ is a subspace of $\text{T}_m(M)^\perp$, and this holds because $\text{T}_m(M)\subseteq \text{T}_m(M_i)$ inside of $\text{T}_m(N)$ $($as the inclusion on $M$ into $N$ factors through the inclusion of $M_i$ into $N)$.
Since each $j_i^*((TM_i)^\perp)$ is a subbundle of $(TM)^\perp$, we get a summation mapping$$\bigoplus_i j_i^*((TM_i)^\perp) \to (TM)^\perp$$that on fibers is the natural map $\bigoplus\text{T}_m(M_i)^\perp \to \text{T}_m(M)^\perp$ using addition in $\text{T}_m(N)$. The natural maps $(TM)^\perp \to \text{N}_{M/N}$ and $(TM_i)^\perp \to \text{N}_{M_i/N}$ are isomorphisms. Pulling back the latter along $j_i$ we get isomorphisms $j_i^*((TM_i)^\perp) \cong j_i^*(\text{N}_{M_i/N})$, and the diagram$$
\require{AMScd}
\begin{CD}
\bigoplus_i j_i^*((TM_i)^*) @>>> (TM)^\perp \\
@V\cong VV     @V \cong VV  \\
\bigoplus_i j_i^*(\text{N}_{M_i/N}) @>>> \text{N}_{M/N}
\end{CD}
$$
with vertical isomorphisms commutes $($check on fibers$)$. Since the bottom side was shown to be an isomorphism above, the top side must be an isomorphism. The meaning of the isomorphism is that motion away from $M$ in orthogonal directions decomposes into a sum of orthogonal motions that are each perpendicular to exactly one of $M_i$ and contained in $\bigcap_{i' \neq i} M_{i'}$. This is easily visualized by considering two surfaces in $\mathbb{R}^3$ that meet transversally along a curve. $($Draw the picture with tangent planes not meeting in a perpendicular manner in order to keep the visualization sufficiently generic.$)$ 
