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In our lecture we defined a category $\mathcal C$ to have arbitrary products, iff every diagram $A:\mathcal J \to \mathcal C$ with $\text{Morph}(\mathcal J)=\{\text{id}_j\}_{j \in Ob(\mathcal J)}$ has a limit. We showed that the category Man (with manifolds modelled on locally convex spaces) possesses finite products (i.e. $Ob(\mathcal J)$ finite), but argued that it does not possess infinite ones. As far as i can see, in defiance of this in the following we always use the product over $Ob(\mathcal J) = \mathbb N$.

My question is now: Is the infinite (countable) cartesian product of manifolds still a manifold? If not, what is the problem? Does the category of locally convex spaces Lcs possess arbitrary products?

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Is your question about the existence of countable products in $\mathsf{Man}$, or is it about the creation of countable products of the forgetful functor to $\mathsf{Top}$? The latter is easy to disprove. – Martin Brandenburg Sep 23 '13 at 10:02
It's about countable products in Man, even though i don't understand what you mean with creation of products of the forgetful functor. – user83496 Sep 23 '13 at 10:12
Well, the question "is the product of manifolds a manifold" doesn't really make sense, since products in an arbitrary category - by definition - are objects of that category. The question is only: do they exist? But usually when people ask something like this, they actually mean something stronger, namely if the products exist and the underlying space is just the usual product of the spaces. So the question is: If we have a countable family of manifolds, can we endow the product of the underlying spaces with the structure of a manifold? Probably almost never. – Martin Brandenburg Sep 23 '13 at 21:53
@MartinBrandenburg I suspect the OP is asking about possibly-infinite-dimensional manifolds, given the mention of "locally convex spaces". – Zhen Lin Sep 23 '13 at 22:28
Hmm, i think it is possible for (cartesian) products of manifolds modelled on normed vector spaces. Take for example $\prod_{\mathbb N} \mathbb S^1$. It is clear that $\mathbb R^{\mathbb N}$ is a locally convex space due to the point-seperating family of semi-norms $p_n((x_i)_{i\in \mathbb N})=|x_n|$. Then one can define the charts canonically, right? – user83496 Sep 23 '13 at 22:30

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