# Subbundle of Hom bundle

I am reading this lecture notes in order to give an alternative proof of the classification of real vector bundles, stated but not proved in Boot and Tu Differential Forms in Algebraic Topology (they reference to Stenrood's book). In page 23 it is stated the following:

Construction 5.7. Let E → M be a vector bundle. Then $hom(E,\mathbb{R}^k)$ is a vector bundle with fibre $hom(E_x,\mathbb{R}^k)$ over $x$. [...] We will also need its subbundle $hom_{inj}(E, \mathbb{R}^k)$ consisting of all injective homomorphisms.

I have tried to check that this is in fact a subbundle (I assumed it reefers to vector subbundle). However, I am having trouble since in order to prove that $hom_{inj}(E, \mathbb{R}^k)$ is a vector subbundle of $hom(E,\mathbb{R}^k)$ I need to prove that the injective homomorphisms between $E_x$ and $\mathbb{R}^k$ are a subspace of constant rank of $hom(E_x,\mathbb{R}^k)$. First of all, the zero homomorphism is not there so...it seems is not a vector subspace. So, my questions are:

• What am I missing?
• It only has the structure of a fiber subbundle but not the structure of a vector subbundle?

Indeed you are right, this is not a vector bundle. But I think it's a fiber bundle. For see this, you can take the map $\pi : Hom(E_x, \mathbb R^k) \to \mathbb R^N, f \to (m_1(f), \dots, m_N(f))$ where $m_i(f)$ are the minors of $f$ of size $l\times l$, where $l = \dim E_x$ . Now, your fiber bundle is exactly $\pi^{-1}(\mathbb R^N \backslash \{0\})$. You can try to use Ehresmann fibration theorem, i.e show that $\pi$ is a submersion, and then it will follows that it is a locally trivial fiber bundle.
• Thanks for the idea! That made think about the following: Could I identify the sets $Hom_{inj}(E_x,\mathbb{R}^k)$ and $Hom_{inj}(\mathbb{R}^n,\mathbb{R}^k)$ where $n$ is the dimension of $E_x$? Because if that where the case, then $Hom_{inj}(E,\mathbb{R}^k)$ would be automatically a fiber bundle.... – D1811994 Apr 20 '17 at 15:17
• You can identify fiberwise $Hom_{inj}$ and $Hom(\mathbb R^n, \mathbb R^k$) for sure but a map $X \to Y$ where all the fibers are isomorphic is not necessary a locally trivial fiber bundle. You also need the local triviality condition. – user171326 Apr 20 '17 at 15:25
I think Ehresmann is a bit of an overkill here—it isn't too hard to check directly that it is a locally trivial fibre bundle with typical fibre the space of injective linear maps $\mathrm{Hom}_{\text{inj.}}(\mathbb{R}^k,\mathbb{R}^n)$ (which itself is not a vector space, at least unless $k$ is $0$).
If $\varphi\colon U\times \mathbb{R}^k\to E|_U$, $(x,v)\mapsto \varphi_x(v)\in E_x$, is a local trivialisation, we consider the induced local trivialisation $\psi\colon U\times\mathrm{Hom}(\mathbb{R}^k,\mathbb{R}^n)\to \mathrm{Hom}(E,\mathbb{R}^n)|_U$, mapping a pair $(x,f)$ to the homomorphism $f\circ \varphi_x^{-1}\in \mathrm{Hom}(E_x,\mathbb{R}^n) = \mathrm{Hom}(E,\mathbb{R}^n)_x$. (This is how the vector bundle structure on $\mathrm{Hom}(E,\mathbb{R}^n)$ is defined.) To conclude what we aim to show, all we have to do is to observe that $\psi$ restricts to an isomorphism $U\times\mathrm{Hom}_{\text{inj.}}(\mathbb{R}^k,\mathbb{R}^n)\to \mathrm{Hom}_{\text{inj.}}(E,\mathbb{R}^n)|_U$.