Let $f: N \to M$ be a smooth immersion and let $p \in M$, $W = f(V) \subset M$ be an submanifold with $q = f(p).$

Then the sequence is split exact

$$T_qW \hookrightarrow T_qM \stackrel{\mu}\to T_qM/T_qW$$

Therefore $T_qM \cong T_qW \oplus T_qM/T_qW.$ Now apparently the quotient according to wikipedia on normal bundles, it can be identified with the normal space (as a fiber) $N_qW.$ That is $NM \stackrel{\pi}\to M$, $$N_qW = T_qM/T_qW = \pi^{-1}(p).$$

I don't know how this last conclusion is reached...

The only answer I am thinking of is that there is some surjective linear (canonical?) map $\ell: T_qW \to N_qW$, then it would yield the commutative diagram:

$$\begin{array}{cccccccc} T_qM & \xrightarrow{\mu} & T_qM/T_qW & \\ \downarrow & \swarrow \\ N_qW \end{array}$$

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    $\begingroup$ Do you know the answer for one point? That is, if $L $ is a subspace of $\mathbb R^n$, how to identify $L^\perp$ to $\mathbb R^n /L$? $\endgroup$ Mar 24, 2019 at 3:52
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    $\begingroup$ @ArcticChar, yes let's just reduce it to that. I actually don't know. The most I could write is $\{r + L: r\in \mathbb{R}^n \} \to \{ g(n,\ell) = 0, \ell \in L \}$. $\endgroup$
    – Lemon
    Mar 24, 2019 at 4:03

1 Answer 1


Actually I just realized what is really going on, the answer is almost silly now that I realize it.

Since $T_qW \subset T_qM$, we can talk about its orthogonal complement $N_q(W)$ which has direct sum with $T_qW \oplus N_qW \approx T_qW \oplus T_qM/T_qV.$ By comparing the elements, it is not surprising that there is an identification between the spaces in question. I just don't know what it is, but my initial question appears to be answered.


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