A linear equation $AX=b$ is solvable if and only if we have that $\mathrm{rk}(A|b) = \mathrm{rk}(A)$. My question is what can we say about the continuity (smoothness?) of the solution?
Precisely, suppose we are given two smooth maps $x\mapsto A_x$ and $x\mapsto b_x$ over some manifold $M$, along with the fact $\mathrm{rk}\big(A_x | b_x\big) = \mathrm{rk}\big(A_x\big)$ at each point $x\in M$. Can we then assert that $AX=b$ admits a smooth solution $X=X(x)$?
I can also pose the same question in a vector bundle setup. Suppose we have a vector bundle map $A:E\to F$ and we are given some section $b\in\Gamma F$ so that $b|_x\in\mathrm{im} A|_x$ for each $x\in M$. Then can we find out a section $X\in \Gamma E$ so that $AX=b$? Note that we can always do this if $A$ is a bundle epimorphism (or even constant rank).
An Attempt : I thought of a possible solution. Introduce a metric on the bundle $E$. Now by the hypothesis, $A_x^{-1}(b_x)\subset E_x$ is a non-empty affine subspace. Hence, there exists a unique vector $X_x$ in the subspace, where the norm $|X_x|$ attains its minimum. Clearly $AX = b$. But I am not sure whether the assignment $x\mapsto X_x$ is continuous/smooth.
Any help regarding this is appreciated. Cheers!