If I have a pullback of $C^*$-algebras and I consider the underlying banach spaces Banach, in particular all maps in the pullback square are contractive, and I have a contractive split $s$ for one map, say $f$. You immediately obtain a split $s'$ for the pullback $f'$ of $f$, by the universal property. But why is the split $s'$ also contractive? Or is this even false?
2026-02-24 07:35:35.1771918535
Is the pullback of a contractive split again contractive?
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The pullback is given by the restricted direct sum with maximum norm. Without loss of generality the pullback of $f$ is then given by the projection onto the first summand. Clearly this admits a contractive split.