This is a cross-post to the [question](https://math.stackexchange.com/questions/2816308/gluing-locally-defined-continous-functions-over-complex-domain) I asked at MSE over almost a month ago.


Suppose $n, l, m \in \mathbb N$ and $n \ge l > m$. Let $T: \mathbb C \to \mathcal M(n \times l; \mathbb C)$ be continuous and $T(\lambda)$ has constant rank $m$ for every $\lambda \in \mathbb C$. Suppose for every $\lambda \in \mathbb C$, there is an open neighborhood $U(\lambda)$ of $\lambda$ and a locally defined continuous function $h\_{\lambda}: U(\lambda) \to \mathcal M(n \times m; \mathbb C)$ where the image $h\_{\lambda}(\beta)$ is a basis for $T(\beta)$ for every point $\beta \in U(\lambda)$. Could we construct a globally continuous function $\phi: \mathbb C \to \mathcal M(n \times m; \mathbb C)$ by gluing together these locally defined $h\_{\lambda}$'s, such that
 the image $\phi(x)$, of every point $x \in \mathbb C$, is a basis for $T(x)$? 


What I have in mind: Suppose we have two continuous families of $g\_1: \mathbb C \supseteq U\_1 \to \mathcal M(n \times m; \mathbb C)$ and $g\_2 : \mathbb C \supseteq U\_2 \to \mathcal M(n \times m; \mathbb C)$ where $U\_1$ and $U\_2$ are two open disks in $\mathbb C$ with $U\_1 \cap U\_2 \neq \emptyset$. Then $(g\_1^1(x), \dots, g\_1^m(x))$ is a basis for $T(x)$ for all $x \in U\_1$ and $(g\_2^1(y), \dots, g\_2^m(y))$ is a basis for $T(y)$ for all $y \in U\_2$ where $g\_i^j: x \mapsto \mathbf c\_j( g\_i(x))$ where $\mathbf c\_j(\cdot)$ denotes the operation of taking the $j^{th}$ column of a matrix. Now for every $y \in U\_2 \cap U\_1$, ${g}\_1 (y) = g\_2(y) S(y)$ for some $S(y) \in GL\_m(\mathbb C)$. It follows $S(y) = (g\_2^T(y) g\_2(y))^{-1} g\_2^T(y) g\_1(y)$. By Cramer's rule, $S(y)$ is continuous with respect to $y$. That is we have a well defined continuous function on $U\_1 \cap U\_2$, $S \colon U\_1 \cap U\_2 \to GL\_m(\mathbb C)$. Since $U\_1 \cap U\_2$ is simply connected, there is a continuous retract $r :\mathbb C \to U\_1 \cap U\_2$. It follows $S \circ r : \mathbb C \to GL\_m(\mathbb C)$ is well defined and continuous with $S \circ r|\_{U\_1 \cap U\_2} = S$. So we can define $g$ by
 \begin{align\*}
 g(x) =
 \begin{cases}
 g\_1(x), & \text{ if } x \in U\_1 \setminus U\_2 \\
 g\_2(x)(S\circ r)(x) , & \text{ if } x \in U\_2.
 \end{cases}
 \end{align\*}
 This function is clearly continuous by construction and satisfies the prescribed condition.


I am not sure whether above argument is completely correct. Even if it was correct, my problem to proceed is: after we glue some collection of maps, the domain $U$ would become not so "regular". I am not sure the intersection of $U$ and an open disk would still be simple connected.



Gluing locally defined continous functions over complex domain
I am not sure whether this gluing method will work, but as I mentioned in the comment, this problem is equivalent to "whether every rank $m$ (typo in the comment, sorry) complex vector bundle over $\mathbb C$ is trivial". To see this, first construct the trivial bundle of rank $n$ over $\mathbb C$, that is, $\mathbb C\times\mathbb C^n$. For every $x\in\mathbb C$, we have an $m$ dimensional subspace $E\_x$ spanned by the columns of the matrix $T(x)$. By the local condition, there is a map $\phi\_x:U(x)\times\mathbb C^m\to\cup\_{y\in U(x)} E\_y$, $(y,v)\mapsto h\_x(y)v$. The transition maps $\phi\_y^{-1}\circ\phi\_x: (U(x)\cap U(y))\times\mathbb C^m\to(U(x)\cap U(y))\times\mathbb C^m$ are continuous, as the second paragraph of the question shows. Therefore $\phi\_x^{-1}$ are local trivializations of $E:=\cup\_{x\in\mathbb C} E\_x$, which is then a complex vector bundle of rank $m$ over $\mathbb C$. The desired global continuous function $\phi$ similarly gives rise to a continuous function $\mathbb C\times\mathbb C^m\to E$ mapping $\{x\}\times\mathbb C^m$ to $E\_x$, and hence establishes an isomorphism between $E$ and the trivial bundle $\mathbb C\times\mathbb C^m$.


Now the question reduces to "whether every rank $m$ complex vector bundle over $\mathbb C$ is trivial". The answer is **Yes** and the reasoning is the following. Since $\mathbb C$ is contractible, the identity map $i$ on $\mathbb C$ is homotopic to the map $j:\mathbb C\to\{0\}$. By Theorem 1.6 of <http://pi.math.cornell.edu/~hatcher/VBKT/VB.pdf>, $E=i^\*E$ is isomorphic to $j^\*E=\mathbb C\times E\_0$, which is trivial because $E\_0$ is isomorphic to $\mathbb C^m$.


**PS** The OP asks the question in the continuous category (on which this answer is based) but in the comments the OP mentions the smooth category. It's an exercise left to the reader (and the OP) to carry the above argument over to the smooth category.


**PPS** The cited book (by [Allen Hatcher](https://en.wikipedia.org/wiki/Allen_Hatcher), who also happens to be a MO user) is a wonderful introduction to vector bundles (and beyond).


