Making a diagram commute Suppose we have a $k$-vector space $A$, two linear maps $$T_1, T_2:A\to A$$ and a bilinear map $\mu:A\otimes A\to A$.
Is there a way to explicitly construct a map $F$ such that $$F\circ \mu = \mu\circ (T_1\otimes T_2)$$
i.e. a way to make a commutative square. In particular, can we construct such an F so that $ker F \subset \mu (ker (T_1\otimes T_2))$?

You're given a "multiplication", and you're asking whether for any two linear maps $T_1,T_2$, $T_1(x)T_2(y)$ can be expressed linearly in terms of $xy$. 
That's not even true for some really nice $\mu$'s, like ones coming from $k$-algebras. 
Take, e.g. $\mathbb C$ as an $\mathbb R$-algebra, it has the usual $\mu : \mathbb C\otimes_\mathbb R\mathbb C\to \mathbb C$. However, something such as complex conjugation makes what you want impossible : take $T_1 =id_\mathbb C$, $T_2=$ complex conjugation, then you're asking if $(x,y)\mapsto x\overline y$ is ($\mathbb R-$)linear in $xy$. 
That is clearly not possible : in fact the first map is not even a function of the second one : take $y=\overline x$, then the second map yields $|x|^2$, that would mean that $x^2$ is a function of $|x|^2$, which is not true. 
