I've been studying some category theory lately and in particular, I became acquainted with the notions of products and coproducts, which led me to ponder the following:

Consider the category of all complex Hilbert spaces (the morphisms being linear isometries). This category has coproducts, due to the direct sum construction: if $X_{\alpha} , {\alpha\in\Lambda}$ is family of Hilbert spaces, define $X := \bigoplus_{\alpha\in\Lambda}X_{\alpha}$ as the set of all "$\Lambda$-tuples" $(x_\alpha)_{\alpha \in \Lambda}$ such that:

$x_\alpha \in X_\alpha \\ \forall \alpha$ and $\sum_{\alpha \in \Lambda} \|x_{\alpha}\|^2 < \infty$

Then one can define addition, scalar multiplication and an inner product on $X$ in an obvious way, and we have the canonical inclusion maps.

However, I don't see any way to make this construction into a product, though maybe there is another construction I don't know of.

I'm sorry if this question is elementary for category theorists, but to me it's not so obvious.

**EDIT**: Thanks for the replies. As it was pointed out, this category doesn't even admit finite products with morphisms being linear isometries. As I don't see any more natural choice for morphisms, I suppose there isn't any good answer to my question (other than "no" :)).

`$X_\alpha$`

's into some Hilbert space $Y$ by linear isometries whose ranges are not orthogonal; how would those give a morphism from $X$ to $Y$? For the simplest case, consider the sum of two copies of the same $X$ and try to form the co-diagonal map to $X$; it won't be an isometry (or even injective). – Andreas Blass Jul 31 '10 at 22:26