Sections form a vector space

Question

Let $p:E \to X$ be a vector bundle and denote by $\Gamma(E)$ the set of all sections of $E$. I want to show that $\Gamma(E)$ is a real vector space and will show what I tried for the sum operation. For sections $s_1,s_2 \in \Gamma(E)$ and $x \in X$ we can define $s_1+s_2(x):=s_1(x)+s_2(x)$ which will give a well defined function $$s_1+s_2:X \to E$$ with the property that for every $x \in X$ we get that $p((s_1+s_2)(x))=x$ since the sum $s_1(x)+s_2(x)$ is the sum in $p^{-1}(x)$. So the only thing that one needs to show is that the sum is continuous.

We know that for every $x \in X$ there is an open $U$ around $x$ and an isomorphism of vector bundles $f:p^{-1}(U) \to U \times \mathbb{R}^n$ for some $n$. Thus we get a function $$U \to p^{-1}(U) \to U \times \mathbb{R}^n, u \mapsto s(u) \mapsto f(s(u)).$$ Note that on ever fiber $p^{-1}(u)$ with $u \in U$ there is a unique homomorphism $g_u:p^{-1}(u) \to \mathbb{R}^n$ with $f(v)=(u,g_u(v))$ where $v \in p^{-1}(u)$. So the composition above is given by $u \mapsto (u,g_u(s(u))$. In particular we get that the function for $s_1+s_2$ is given by: $$u \mapsto (u,g_u(s_1(u)+s_2(u))).$$ Defining the function $g:p^{-1}(U), v \mapsto g_u(v)$, $p(v)=u$ we get that $g$ is continuous because $g=\pi_2 \circ f$ where $\pi_2$ is the projection onto the second coordinate. We also know that the sum is continuous, and so the function $u \mapsto (u,g_u(s_1(u)+s_2(u)))$ is continuous. Composing this with the homeomorphism $f^{-1}$ we get the function $u \mapsto s_1(u)+s_2(u)$. So using that composition of continuous functions remains continuous we get that the sum $s_1+s_2$ is continuous on $U$. Therefore we can conclude that $s_1+s_2$ is continuous on $X$.

1. Is my approach correct? I think that I made at least one mistake namely when saying that the sum is continuous. While it is continuous on $p^{-1}(u)$ I think its not necessarily continuous on $p^{-1}(U)$ (when switching between fibers) which is what I need for the composition. I think this can be fixed by noting that $g_u$ is linear and so we have that the function $u \mapsto (u,g_u(s_1(u)+s_2(u))$ is equal to $u \mapsto (u,g_u(s_1(u))+g_u(s_2(u)))$ and so we get addition in $\mathbb{R}^n$ which is continuous.

2. I have also read this question where the answer does a similar approach I think. However, it is claimed that it is necessary to show the independence of the trivialization $f$. I don't see why this is the case. What I have shown is that using the existence of some trivialization we get that the sum is continuous, I don't see why one would need to show the independence here.

Answer 1

Your approach is correct. For any vector bundle, you correctly defined the sum of two sections, but concerning continuity of the sum you have indeed a gap.

I suggest to proceed as as follows.

1. Consider a trivial bundle $\pi : Y \times \mathbb R^n \to Y$. Then we have a bijection
   $$b : C(Y,\mathbb R) \to \Gamma(Y \times \mathbb R^n), b(f)(y) = (y, f(y)). $$ Here $C(Y,\mathbb R)$ denotes the vector space of continuous real-valued functions on $Y$. Using this bijection we easily see that the sum $s_1 + s_2$ of sections is again a (continuous) section: We have $b^{-1} (s_1) + b^{-1}(s_2) \in C(Y,\mathbb R)$, thus $b(b^{-1} (s_1) + b^{-1}(s_2)) \in \Gamma(Y \times \mathbb R^n)$. But clearly $b(b^{-1} (s_1) + b^{-1}(s_2)) = s_1 + s_2$. Similarly a scalar multiple of a section is a section.

2. Now consider an arbitrary vector bundle $p : E \to X$. Since continuity is a local property, it suffices to show that $(s_1 + s_2) \mid_U : U \to E$ is continuous for each open $U \subset X$ such that the bundle is trivial over $U$ which means that there exists an isomorphism of vector bundles $f :p^{-1}(U) \to U \times \mathbb R^n$. Thus it suffices to show that $f \circ (s_1 + s_2) \mid_U$ is continuous. We have $f \circ (s_1 + s_2) \mid_U = f \circ s_1 \mid_U + f \circ s_2 \mid_U$, where the $f \circ s_i \mid_U$ are sections of the trivial bundle $U \times \mathbb R^n$. Now apply 1.

Note that the above proof clearly shows that it suffices to consider a single trivialization around each point of $X$. No transition functuons between different trivializations are involved.