I got stuck at the following problem.
Let $X,Y$ be normed spaces. A bounded linear operator $\tau\in\mathcal{B}(X,Y)$ is called strictly coisometric if $ \tau(\operatorname{Ball}_X(0,1))=\operatorname{Ball}_Y(0,1)) $ which is equivalent to that $\Vert\tau\Vert=1$ and for all $y\in Y$ there exist $x\in X$ so that $\tau(x)=y$ and $\Vert x\Vert= \Vert y\Vert$.
Is it true that the functor $\mathcal{B}(Z,-)$ preseve coisometries, i.e. is true that given $\tau$ strictly coisometric operator $ \mathcal{B}(Z,\tau):\mathcal{B}(Z,Y)\to\mathcal{B}(Z,X):\varphi\mapsto\tau\circ\varphi $ is strictly coisometric too.