If I choose a principal bundle, let us say $G\rightarrow P \rightarrow B$, with $G=U(1)$, $P=S^1 \times S^1$ and $B=S^1$. Can I follow the identity element of the group over a curve at the base. How isn't this equivalent to a preferred Horizontal subspace (and hence a canonical connection) ?



Non existence of preferred Horizontal subspace on a bundle
1. A trivial principal bundle, $P= G\times B\to B$ has a preferred connection (which is also flat). There is an obvious map $B\to \{1\}\times B\subset P$ and you can take as horizontal bundle the image of its differential.
2. If your principal bundle $P\to B$ is not trivial, but is trivializable, i.e. exists a section $\sigma:B\to P$, then you can use this section to trivialize it, i.e. construct a isomorphism with the trivial bundle, $\Phi\_\sigma: P\overset{\simeq}{\to} G\times B$. Over $G\times B$ you have the preferred connection of item 1) and you can use $\Phi\_\sigma$ to pull it back to $P$.
You now have a connection on $P$, but the latter depends on the choice of the trivialization $\sigma$. How many choices do you have? Well, the space of sections is acted transitively and freely by (hence is in bijection with) the gauge group $\mathcal G(P) \simeq \operatorname{Maps}(B, G)$ as you can alter $\sigma$ by multiplying it with an element of $\operatorname{Maps}(B, G)$. This tells you that the possible choices that you have are $\mathcal{G}(P)$.
Notice that all these connections will be flat.


3. If the principal bundle $P$ is not trivializable, then you cannot apply any of the above and you have a space of connection which is an *affine* space modelled over $\Omega^1(B; \operatorname{ad} (P))$ where $\operatorname{ad} (P) = P\times\_{\operatorname{ad}}\mathfrak{g} $ is the bundle associated to the adjoint action $\operatorname{ad}:G\to \operatorname{Aut}(\mathfrak{g})$ of $G$ on its Lie algebra $\mathfrak{g}$.
As an example of non-trivializable bundle you can take $P$ equal to the bundle of oriented orthonormal frames of the tangent bundle of $\mathbb{S}^2$ (after you choose an orientation for $\mathbb{S}^2$ of course); this is an $\mathbb{S}^1\simeq \operatorname{SO}(2)$ bundle over $\mathbb{S}^2$ which does not admit any section.


As a side note, even in this case the gauge group $\mathcal{G}(P)$ acts on the set of connections, however it is no more in bijection with $\operatorname{Maps}(B,G)$ but instead with $\Gamma(B; \operatorname{Ad}(P))$ i.e. sections of the bundle associated to the action $\operatorname{Ad}: G\to \operatorname{Aut}(G) $, $\operatorname{Ad}(g)\cdot p = gpg^{-1}$.
In the case of a trivial bundle $\operatorname{Ad}(P)\simeq G\times B$ hence in item 2) we were able to identify $\mathcal{G}(P)$ with $\operatorname{Maps}(B, G)$.


