@NoamD.Elkies: Yes, that's what I mean by $C_0(\mathbb{R})$.

@Noam: For $G=\mathbb{R}$, my question comes down to: If $f\in C_0(\mathbb{R})$ is the Fourier transform of some function in $L^1(\mathbb{R})$, is then $|f|$ the Fourier transform of some other function in $L^1(\mathbb{R})$?

Yes, for every locally compact abelian group $G$, the Fourier algebra $A(G)$ is naturally isometrically isomorphic to $L^1(\widehat{G})$ via the Fourier transform. In particular, since $\widehat{\mathbb{R}}\cong\mathbb{R}$, we have $A(\mathbb{R})=\{\widehat{f} : f\in L^1(\mathbb{R})\}$, with the norm coming from $L^1(\mathbb{R})$.

So let $y\in Y$. Since $\varphi$ is surjective, there exists $x\in X$ such that $y=\varphi(x)$. By assumption, there exist $x_0$ and $c_k$ such that $x=x_0+\sum_k c_k x_k$. Using that $\varphi$ is linear, it follows $y=\varphi(x)=\varphi(x_0)+\sum_k c_k y_k$. Since $\varphi(x_0)\in Y_0$, the claim is proved.

More generally: If $\varphi\colon X\to Y$ is a surjective linear map between vector spaces, and $X_0$ is a subspace of $X$ of codimension $n$, then $Y_0:=\varphi(X_0)$ is a subspace of $Y$ of codimension at most $n$. Indeed, let $x_1,\ldots,x_n\in X$ such that every $x\in X$ can be written as $x=x_0+\sum_k c_k x_k$, for $x_0\in X_0$ and coefficients $c_k$. We claim that the analogous statement holds in $Y$, with $Y_0$ and the vectors $y_k:=\varphi(x_k)$, $k=1,\ldots,n$. (This will show that $Y_0$ has codimension at most $n$ in $Y$.)

Yes, I agree. But the homomorphism $C([0,1],M_n)\to C(\mathbb{T},M_n)$ feels to me 'more natural' than the map $C(\mathbb{T},M_n)\to C([0,1],M_n)$. For instance, the latter 'forgets' K-theory.

Ah I see. Thanks. But then what is "the cocycle" associated to the cross section? Maybe there are two. One measuring if the extension is split, and one measuring if the section is multiplicative.

Really? The usual cocycle for the section $\sigma$ is the map $\omega\colon G\times G/N\to N$, defined by the formula $\sigma(gy)\omega(g,y)=g\sigma(y)$. If $\sigma$ is multiplicative (which means that $G$ is a semidirect product), then $\alpha$ is trivial, but the cocycle $\omega$ need not be. How can this be?

Do you know if there is also a counterexample to the modified question of the comment above?

Dear Dave, thank you for your answer. The reason I was thinking that semidirect products work is that I thought the cocycle records how far $\sigma(xy)$ is from $\sigma(x)\sigma(y)$ for $x,y\in G/N$. I understand now that the cocycle is in fact not recording this. Let us therefore consider $\alpha\colon G/N\times G/N\to N$ defined by the formula $\sigma(xy)\alpha(x,y)=\sigma(x)\sigma(y)$. Then, I should have asked my question for $\alpha$: Given a compact subset $K$ of $G/N$, can $\sigma$ be chosen s.t. $\alpha(K\times G/N)$ is pre-compact. Then the answer is 'yes' for semidirect products.