# Strict coisometries and operator norm.

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.

-

Your definition of strict coisometry implies surjectivity but not injectivity. Take advantage of this. There could be a subspace of $X$ isometric to $Y$ and another subspace of $X$ not isometric to $Y$, also mapped to $Y$ by $\tau$ but where the norm is not preserved.
Further details: Fix $0<\alpha<1$ and $\beta>1$ such that $\alpha\beta<1$.
Let $X=\mathbb C \oplus \mathbb C$ with the norm $\Vert(x_1,x_2)\Vert=\max\{|x_1|,\beta|x_2| \}$ and $Z=Y=\mathbb C$ with standard norms. Then $\tau:X \to Y:(x_1,x_2)\mapsto x_1+x_2$ is strictly coisometric.
Define $\varphi:Z\to Y:z\mapsto\alpha z$ and $\psi:Z\to X: z\to(0,\alpha z)$. Then $\varphi=\tau \circ \psi$, but we have $\Vert\varphi\Vert=\alpha$ and $\Vert\psi\Vert=\alpha\beta>\alpha=\Vert\varphi\Vert$.