Let $\mathsf{Hilb}_1$ (resp. $\mathsf{Hilb}_{\leq 1}$) denote the category of Hilbert spaces with linear isometries (resp. short linear maps).

Does $\mathsf{Hilb}_1$ (resp. $\mathsf{Hilb}_{\leq 1}$) have binary coproducts?

Notice what is usually called the direct sum of Hilbert spaces is not a coproduct, since here the summands are orthogonal to each other, which is a special relation not suitable for a coproduct. But perhaps there is another construction which turns out to be the coproduct.


This is an answer for $\mathsf{Hilb}_1$. Let $X$ be any nonzero Hilbert space. I claim the coproduct $X \vee X$ doesn't exist in $\mathsf{Hilb}_1$.

Assume that it did, call it $X \xrightarrow{i_0} X \vee X \xleftarrow{i_1} X$. Then by the universal property of the coproduct, we get an isometry $f : X \vee X \xrightarrow{(\operatorname{id}, \operatorname{id})} X$ such that $f \circ i_0 = \operatorname{id}_X$ and $f \circ i_1 = \operatorname{id}_X$. Since $f$ is an isometry and hence injective, it follows that $i_0 = i_1$.

Now consider the isometry $\theta : X \to X$ given by $x \mapsto -x$. Again by the universal property there is an isometry $g : X \vee X \xrightarrow{(\operatorname{id}, \theta)} X$ such that $g \circ i_0 = \operatorname{id}_X$ and $g \circ i_1 = \theta$. But this is absurd, since $\theta$ isn't the identity but $i_0 = i_1$.

  • $\begingroup$ Thank you! Perhaps something similar works for short linear maps. $\endgroup$ – Martin Brandenburg Sep 9 '15 at 9:46
  • 2
    $\begingroup$ What you have shown is this: if you have a category in which every morphism is monic and coproducts exist, then the category is a preorder. $\endgroup$ – Zhen Lin Sep 9 '15 at 10:38

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