Why can we always take the zero section of a vector bundle?

$\require{AMScd}$ As I understand it, a rank $k$ vector bundle is a pair of topological spaces with a map between them $$ E\xrightarrow{p}B $$ such that there exists an open cover $(U_\alpha)$ of $B$ over which $E$ locally looks like $U_\alpha\times\mathbb R^k$. Precisely, we have homeomorphisms $$ \phi_\alpha\colon p^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^k $$ such that the diagram \begin{CD} p^{-1}(U_\alpha) @>\phi_\alpha>> U_\alpha\times\mathbb R^k\\ @VVbV @VV\text{pr}_1V\\ U_\alpha @>\text{id}>> U_\alpha \end{CD} commutes (I can't do diagonal arrows).

A section of the vector bundle is a continuous map $s\colon B\to E$ such that $p\circ s=\text{id}_B$. A basic fact about vector bundles is that every vector bundle admits at least one section, namely the zero section, obtained by identifying each $p^{-1}(U_\alpha)$ with $U_\alpha\times\mathbb R^k$ and taking $s(b)=0$. Explicitly, we are taking: $$ s(b)=\phi_\alpha^{-1}((b,0)) $$ for $b\in U_\alpha$.

But in order to get a well defined continuous map, we need to ensure that $\phi_\alpha^{-1}(b,0)=\phi_\beta^{-1}(b,0)$ if $b\in U_\alpha\cap U_\beta$. Now one thing we certainly can't do is require that $\phi_\alpha$ and $\phi_\beta$ actually agree on the intersection - then we could glue them all together to get a global trivialization of the bundle, which we can't do in general. Why then are we able to insist that they agree at a certain point? To me, it seems equivalent to the existence of a global section, and hence completely circular.

There must be something obvious I'm missing. In the meantime, here's an equivalent formulation of the problem: we define an affine bundle $E\xrightarrow{p}B$ to be the same as a vector bundle, but we identify the fibres with $k$-dimensional affine linear subspaces of some fixed real vector spaces. So they needn't contain a special zero point.

Since affine spaces are homeomorphic to vector spaces of the same dimension, any rank $k$ affine bundle is isomorphic to some rank $k$ vector bundle (though I'm not at all sure about whether this is true, for similar reasons). So an affine bundle should admit a global section, but there is no easy way to see how to get one.

Solution 1:

In your definition you are missing the condition that the transition maps for a vector bundle are supposed to be linear fiberwise. Once you assume that, the problem you're facing in gluing will go away.

Solution 2:

If you had a canonical zero element in each fiber, you would see the section right? So but that is already true, since you usually require in the definition that your fibers admit a vector space structure and you also require that the local trivializations are isomorphisms i.e. also preserve zero. That gives you a continous choice of zeros everywhere.

Solution 3:

I think it's more straightforward than you're currently thinking. Every fiber $F_b$ has the structure of a vector space, and therefore has a zero element $0_b$. Sending an element $b$ to the zero vector $0_b$ defines the (continuous) zero section you're after.