Our following question is motivated by [this very interesting answer](https://mathoverflow.net/questions/258293/is-a-pointwise-simple-tensor-valued-continuous-map-a-tensor-product-of-two-co/258297#258297)


Assume that $A$ is a $C^{\*}$ algebra. Put $X=\{a\otimes b \mid a,b \in G(A)\}$ where $G(A)$ is the space of all invertible elements of $A$. 


Assume that there are two continuous maps $f,g: X \to A$ such that $ x=f(x)\otimes g(x),\;\;\;\forall x \in X$.



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> Does this imply that $A$ is a finite dimensional commmutative algebra?
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Note: After a few modification the above question can be generalized to the category of Banach spaces or other similar categories.


For example on can consider the following:



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> "Classification of all Banach space $V$ such that for every tensor product norm with $\parallel x\otimes y \parallel= \parallel x \parallel. \parallel y \parallel$ we have two continuous maps $f,g$ from the space of simple tensors to $V$ (or from $V \bar{\otimes} V$ to $V$) with $x=f(x) \otimes g(x)$"
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Continuous factors for invertible simple tensors
Well, you don't get "finite dimensional", since this is true for all infinite-dimensional commutative $C^\*$-algebras too.


We can suppose $A = C\_0(\Omega)$ for some locally compact Hausdorff $\Omega$. Then $X$ is identified with an appropriate subset of $C\_0(\Omega^2)$: those $x : \Omega^2 \to \mathbb{C}$ which can be written $x(\omega\_1, \omega\_2) = a(\omega\_1) b(\omega\_2)$ where $a,b \in C\_0(\Omega)$ are invertible, i.e. nowhere vanishing.


Fix some $\omega\_0 \in \Omega$ and put
$$f(x)(\omega) = \frac{x(\omega, \omega\_0)}{\sqrt{x(\omega\_0, \omega\_0)}}, \qquad g(x)(\omega) = \frac{x(\omega\_0, \omega)}{\sqrt{x(\omega\_0, \omega\_0)}}.$$
Then $f,g : X \to A$ are continuous, and for any $x \in X$, by writing $x = a \otimes b$, it's easy to check that you have $f(x) \otimes g(x) = x$.


