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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.

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up vote 2 down vote accepted

Taking another look, I see that you only wanted some hints, so ignore the second part of this answer until you have solved this problem yourself.

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 $\Vert\psi\Vert=\alpha\beta>\alpha=\Vert\varphi\Vert$.

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