The answer is 'no'.  For example, let
$$
X = \{\ B\otimes C\ | \det(B)\det(C) = 1\ \}\subset \mathrm{SL}(4,\mathbb{C}).
$$
Then $X$ is isomorphic to the group $\mathrm{SO}(4,\mathbb{C})$ and hence $\pi_1(X)\simeq \mathbb{Z}_2$, while the set
$$
G = \{\ (B, C)\ |\ \det(B)\det(C) = 1\ \}\subset 
\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C})
$$
is a connected Lie group that satisfies $\pi_1(G)\simeq \mathbb{Z}$, with generator given by the subgroup
$$
\{\ (B, B^{-1})\ |\ B\in T\ \}
$$
where $T\subset\mathbb{GL}(2,\mathbb{C})$ is a circle subgroup that generates $\pi_1\bigl(\mathbb{GL}(2,\mathbb{C})\bigr)\simeq \mathbb{Z}$.

Now, the map $\tau:\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C})\to \mathrm{GL}(4,\mathbb{C})$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism, and the preimage under $\tau$ of the space $X$ is the two-component subgroup
$$
G^+ =  \{\ (B, C)\ |\ \det(B)\det(C) = \pm1\ \}\subset 
\mathrm{GL}(2,\mathbb{C})\times\mathrm{GL}(2,\mathbb{C}).
$$
(This is because of the identity $\det(B\otimes C) = \det(B)^2\det(C)^2$.)
Note that $\pi_1(G^+) = \pi_1(G) = \mathbb{Z}_2$.

The map $\tau:G\to X$ given by $\tau(B,C) = B\otimes C$ is a group homomorphism.  If there were a continuous map $\sigma = (g,h):X\to G^+$ that satisfied $\tau\circ\sigma = \mathrm{id}_X$, this would induce a homomorphism $\sigma_*:\pi_1(X)\to\pi_1(G^+)$ that satisfied
$$
\tau_*\circ\sigma_* = \mathrm{id}:\pi_1(X)\to\pi_1(X)\simeq\mathbb{Z}_2.
$$
However, then $\sigma_*$ would be a nontrival homomorphism $\mathbb{Z}_2\to\mathbb{Z}$, and this does not exist.