Norm topology and compact open topology coincide for vector bundles? Let $B$ be a compact Hausdorff base space, $p:E \rightarrow B$ a vector bundle. 


*

*As outlined in Hatcher's, pg 45 we may associate a norm to bundle endomoprhisms, denoted $End(E)$, a norm. Given by $$ |\alpha| = \sup_{z \in B} ||\alpha_z||$$
where $||\alpha_z||$ is the norm of the linear operator $\alpha$ when restricted to a fiber of $z$. $|| \cdot ||$ being a norm induced by a choice of inner product on $E$.  

*$End(E) \subseteq E^E$, hence, it inherits as a subspace, of the compact open topology of mappings. 
Are these topologies the same? Is one coarser than the other?

Though I do not really understand the proof given, I have written down a proof that hopefully explicates user10354138's proof. 
 A: They are equivalent.
A linear map $\mathbb{R}^n\to\mathbb{R}^n$ is determined by its $S^{n-1}\to\mathbb{R}^n$ (similarly for $\mathbb{C}$).  So the compact-open topology of $\operatorname{End}E$ can be taken as coming from $E^\Sigma$, where $\Sigma$ is the (fiber, not vector) subbundle of $E$ consisting of unit vectors.  Then $\Sigma$ is compact.  And $\{v\in E\mid\|v\|<r\}$ is open, so $\{|\alpha|<r\}$ is an open set in the compact-open topology.  Obviously we can also translate, so $\{\alpha\mid |\alpha-\beta|<r\}$ is also open.  This proves one inclusion, and the other direction is obvious.
A: I am really slow at making sense of all this. But I think there are some subtleties.
On the inner product:
In each local trivialization 
$$ p^{-1}(U)\rightarrow U \times \Bbb R^n \quad \quad  (*)$$ 
we take an inner product of $\Bbb R^n$, $b_U(-,-):\Bbb R^n \times \Bbb R^n \rightarrow \Bbb R^n$. Then we take a partition of unity, and define $$b(-,-):=\sum \varphi_Ub_U(-,-)$$ as our inner product. This is one way to define an inner product on $E$ and will be assumed to be. 
Locally, only finitely many supports are nonzero, hence restricting $U$ in $(*)$ above, shows that the set $U \times B(\epsilon)$, $B(\epsilon):=\{x \in \Bbb R^n, b(x,x) < \epsilon \}$ is an open set. 

Norm topology is coarser:
Let $b \in B$, and $b \in U$ a local trivialization, such that $b(-,-)$ as an inner product on $E$ allows us to assume it restricts to an inner product on in
$$h: p^{-1}(U) \rightarrow U \times \Bbb R^n. $$
$b$ is contained in a precompact neighborhood $V \subseteq \bar{V} \subseteq U$ as $B$ is compact Hausdorff. Thus, the preimage of $h^{-1}(\bar{V} \times S^1)$, is a comapct subset of $p^{-1}(U)$, hence that of $E$. 
A subbasis element is of the form $W(K,Z)$, $Z$ being an open subset of $E$. For example, we can let 
$$K= h^{-1}(\bar{V} \times \Bbb R^n), Z= h^{-1}(U \times B(\epsilon))$$ 
Then a subbasis element is 
$$ \{\alpha \in End(E) \, : \, ||\alpha_x|| \le \epsilon \text{ for all } x \in \bar{V} \}. $$ 
Thus, as $B$ is compact, and basis elements are finite intersections subbasis ones, 
$$ \{ \alpha \in End(E) \, : \, ||\alpha_x || \le \epsilon \text{ for all } x \in B \}. $$ 
is thus an open set. Hence, the topology of $End(E)$ as subspace is coarser. 

Compact open topology is coarser: Note that if $K_1 \supseteq K_2, Z_1 \subseteq Z_2$, then $W(K_2 , Z_1) \subseteq W(K_1, Z_1)$. Thus, we maximize $K$ and minimize $Z$, to see that $W(B,U_\epsilon)$ forms a basis, where 
$$ U_\epsilon:= h^{-1}(U \times B_n(\epsilon))$$ 
But this is precisely the norm topology.  
