Normal Bundle Orientable and inclusion null homotopic then M is orientable

Question

I am trying to do the following exercise :

Let $M$ be a submanifold of $N$, both without boundary. Show that if the normal bundle of $M$ in $N$ is orientable and $i: M\rightarrow N$ is null-homotopic then $M$ is orientable.

I am bit lost and have no clue where to start. Since the normal bundle is orientable I guess proving the orientability of $M$ would be equivalent to that of $N$ but I am not sure this helps. Also I have no idea where to use the fact that $i: M\rightarrow N$ is null-homotopic, thought about using degree theory but this is only defined when we are working with orientable manifolds.

So we have a map $H:M\times [0,1] \rightarrow N$ such that $H(0,m)=i(m)$ and $H(1,m)=p$ for some $p\in N$. I have no idea what to do with this affirmation , I have thought about tubular neighborhoods , or doing the pull-back by the homotopy but none of this seems useful.

Any hint to get me started is appreciated , just trying to understand how to tackle this.

Thanks in advance.

Answer 1

If $M$ is a submanifold of $N$ with inclusion $i$, then you have the isomorphism $TN|_M \cong TM \oplus \nu$ where $\nu$ is the normal bundle of $M$ in $N$ and $TN|_M$ is the restriction of $TN$ to $M$. Another way to write this last part is that $TN|_M = i^*(TN)$, the pullback of the tangent bundle of $N$ with respect to the inclusion map.

Now, using the fact that homotopic maps yield isomorphic pullback bundles, since $i \simeq c$ for $c$ a constant map, you have $i^*(TN) \cong c^*(TN)$ and hence $c^*(TN) \cong TM \oplus \nu$. From here, you should try to show that $c^*(TN)$ is orientable (what's the pullback of a bundle by a constant map?) and that this together with the fact that $\nu$ is orientable will force $TM$ to be an orientable bundle, which is equivalent to orientability of $M$.

Answer 2

Hint: given a vector bundle $E \to X$ show that there is a corresponding group homomorphism $w_E: \pi_1 X \to \Bbb Z/2$ so that $w_E(\gamma) = 0$ iff $\gamma^*E$ is the trivial bundle over the circle (equivalently, iff $\gamma^*E$ is orientable).

Then show that one has $w_{E \oplus F} = w_E + w_F$, and $E$ is orientable iff $w_E = 0$. From here, determine how to conclude.