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NOTICE : I deleted my previous version of this question because there were some bad typos.

I'm reading some notes online about transversality. For the context, here let $M$ be a (orientable) manifold. Let $S_1$ and $S_2$ be submanifolds of $M$. By definition, an orientation on $S_1$ is an orientation of its tangent bundle $TS_1$. Define $\mathcal{M}:=S_1\cap S_2$. Here's my question: assuming the intersection $S_1\cap S_2$ is a transverse one, the author of the notes claims that the transversality condition implies that the tangent bundle $TS_1$ splits along $\mathcal{M}$ (the transverse subspace of $M$) as

$TS_1|_{\mathcal{M}}\cong T\mathcal{M}\oplus NS_2|_\mathcal{M},$

where $NS_2$ is the normal bundle of $S_2$ and where $\cdot|_{\mathcal{M}}$ denotes the restriction of "$\cdot$" to $\mathcal{M}$. Why transversality implies that? From what I know of transversality -- i.e.

$S_1\pitchfork S_2 \Rightarrow\forall \xi \in (S_1\cap S_2), \mathcal{T}_\xi S_1+\mathcal{T}_\xi S_2=\mathcal{T}_\xi M.$

I cannot see why the direct sum relation holds.

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  • $\begingroup$ Maybe should I use the splitting lemma defining some maps before? $\endgroup$ – DaveWasHere Jul 9 '18 at 20:08

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