This question is basically an extension of a previous question by replacing a matrix with a continuous matrix-valued function: Given a matrix $A \in \mathbb{R}^{m \times n}$ ($m > n$), can one always find an injective matrix $B$ close to $A$?
Suppose that $X \subset \mathbb{R}^{n}$ is compact and $U \subset \mathbb{R}^{m}$ is convex compact. A continuous matrix-valued function $A: X \to \mathbb{R}^{l \times m}$ is given with $l > m$. Given $\epsilon > 0$, can one always find an injective-matrix-valued continuous function $B: X \to \mathbb{R}^{l \times m}$ such that for any $x \in X$, $\lVert A(x)-B(x) \rVert < \epsilon$ and $u_{1} \neq u_{2}$ $\Rightarrow$ $B(x)u_{1} \neq B(x)u_{2}$?
Note from the comments of the previous question that one can construct an injective matrix $B$ for a given matrix $A$ by singular value decomposition. My understanding is something like this:
SVD of $A \in \mathbb{R}^{l \times m}$; $A = U \Sigma V^{*}$. Define $B := U \left( \Sigma + t\begin{bmatrix} I \\ 0 \end{bmatrix} \right) V^{*}$, which satisfies the requirement of the previous question with arbitrarily small $t \in \mathbb{R}_{>0}$. However, this approach is not easily extended for this question as the SVD of $A(x)$ is not uniquely determined in general for given $x \in X$ and it seems hard to guarantee that there exists $U(x)$, $V^{*}(x)$, and $\Sigma(x)$ continuous w.r.t. $x$.
Edit: The following simple example should work: Consider $X \subseteq \mathbb R^2$ to be the closed unit ball. Now the $2\times 1$ matrix $A(x) = x$ cannot be perturbed to avoid $(0,0)$ while staying continuous.
Consider $X\subseteq\mathbb R^3$ to be the unit sphere. $$A(x) = \begin{bmatrix} 1&\\0&x\\0&\end{bmatrix}.$$ Any small enough perturbation cannot change the homotopy class of this map, so the second column (which is never zero when $\epsilon < 1$) after normalizing will have mapping degree one. Similarly the first column will have mapping degree zero. A standard topological argument shows that these two normalized vectors will be equal at some point, and therefore the two actual vectors will be colinear, rendering $B$ non-injective.
(I haven't thoroughly checked this, but something along the line will surely work.)