Let $A$, $B$ are two Banach space, on the algebraic tensor space $A$ $\odot$ $B$, we can define the projection(maximal) tensor norm $\gamma$ and injective(minimal) tensor norm $\lambda$. For the algebraic tensor space $A*$$\odot$$B*$, we have $\gamma*$=$\lambda$, but $\lambda$* maybe is not $\gamma$.
How to explain that the dual of injective tensor norm maybe is not projective tensor norm? Any simple counterexample?
We claim that $$ (c_0\otimes_\varepsilon c_0)^*\neq c_0\otimes_\pi c_0 $$ From this answer we know that $c_0\otimes_\varepsilon c_0=c_0$, so $(c_0\otimes_\varepsilon c_0)^*=c_0^*=\ell_1$ By result of Kothe the only infinite dimensional complemented subspce of $l_1$ is $l_1$.
Now, for any Banach spaces $X$, $Y$ the space $X$ is complemened in $X\otimes_\pi Y$ (see ex 2.1 in Introduction to tensor products of Banach spaces. R. A. Ryan). Thus $c_0\otimes_\pi c_0 $ have complemented subspace isomorphic to $c_0$.
Since $c_0$ is not isomorphic to $\ell_1$ ($c_0$ is not a dual space while $\ell_1=c_0^*$), then $(c_0\otimes_\varepsilon c_0)^*$ is not isomorphic to $c_0\otimes_\pi c_0$.