I'm curious about which (categorical) products exist in the categories of topological manifolds with continuous mappings/smooth manifolds with smooth mappings. Here is what I have so far:

Finite products exist in both categories and are given by the cartesian product of the underlying sets equipped with the product topology. Products of Hausdorff spaces are Hausdorff, at most countable products of second-countable spaces are second-countable and charts on the product are given by products of charts on the factors. In the smooth case, the products of charts from the respective $C^{\infty}$-atlases form again a $C^{\infty}$-atlas.

Lemma: If $M,N$ are top. manifolds of dimension $m,n$ respectively and $m>n$, there is no continuous injection $M\rightarrow N$.

Proof: Assume $f\colon M\rightarrow N$ is a continuous injection. Pick $p\in M$ and pick a chart $\psi\colon V\rightarrow\psi(V)$ of $N$ about $f(p)$. Then $f^{-1}(V)$ is open, hence contains a chart domain $U$ about $p$ with corresponding chart $\varphi\colon U\rightarrow\varphi(U)$ of $M$. Then $\psi\circ f\circ\varphi^{-1}$ is a continuous injection from $\varphi(U)$ into $\psi(V)$, but after post-composing with the inclusion $\psi(V)\subseteq\mathbb{R}^n\subseteq\mathbb{R}^m$, invariance of domain implies that $\psi(V)\times\{0\}^{m-n}$ is open in $\mathbb{R}^m$, which is a contradiction.

Next, I claim that if $(M_i)_{i\in I}$ is an infinite collection of manifolds, infinitely many of which have positive dimension (and none of which are empty), then their product doesn't exist. Suppose, to the contrary, that $(P,(\pi_i)_{i\in I})$ is their product. Let $J\subseteq I$ be a finite subset and consider the product (which exists by the preceding paragraph) $(\prod_{j\in J}M_j,(p_j\colon\prod_{j\in J}M_j\rightarrow M_j)_{j\in J})$. Pick arbitrary morphisms $p_i\colon\prod_{j\in J}M_j\rightarrow M_i$ for all $i\in I\setminus J$ (constant ones do the job in either category, since $M_i$ is non-empty). Together, these factor as a morphism $f\colon\prod_{j\in J}M_j\rightarrow P$ satisfying $\pi_j\circ f=p_j$ for all $j\in J$. Consider an arbitrary object $C$ and morphisms $r,s\colon C\rightarrow\prod_{j\in J}M_j$ such that $f\circ r=f\circ s$. Then, for every $j\in J$, $p_j\circ r=\pi_j\circ f\circ r=\pi_j\circ f\circ s=p_j\circ s$, so, by a property of products, $r=s$. That is, $f$ is a monomorphisms, hence injective (for the same reason as in $\mathbf{Set}$, since singletons are manifolds). The Lemma now implies $\sum_{j\in J}\dim(M_j)=\dim(\prod_{j\in J}M_j)\le\dim(P)$ for all finite subsets $J\subseteq I$, but since there are infinitely many $M_i,i\in I$ with positive dimension, this is impossible.

On the other hand, if $(M_i)_{i\in I}$ is an infinite collection of manifolds and all but finitely many of the $M_i,i\in I$ are singletons, I claim their product exists. More precisely, if $M_{i_1},...,M_{i_n}$ are the finitely many non-singletons among the $M_i,i\in I$, I claim the product is given by $\prod_{j=1}^nM_{i_j}$ with the usual projections $p_j\colon\prod_{j=1}^nM_{i_j}\rightarrow M_{i_j}$ for $j=1,...,n$ and the unique maps $\prod_{j=1}^nM_{i_j}\rightarrow M_i$ for $i\in I\setminus \{i_1,...,i_n\}$. This is essentially obvious, because singletons are terminal in either category.

Question: First of all, is all of the above correct? Second of all, what can we say about the existence of a product for an infinite family of manifolds, only finitely many of which have positive dimension and infinitely many of which are not singletons, in either category?

Edit: I had forgotten and just recalled that we usually do consider the empty set a manifold. The empty set is initial in either category. If a collection $(M_i)_{i\in I}$ contains the empty manifold, its product exists and is given by the empty set with the only possible maps for the obvious reasons. My above argument for non-existence fails in case one of the $M_i,i\in I$ is empty at the stage of picking an arbitrary morphism into $M_i$.

Edit 2: Here is an argument for the remaining case: Let $(M_i)_{i\in I}$ be a collection of manifolds, none of which are empty, only finitely many of which have positive dimension and for which not all but finitely many of them are singletons. Assume the product $(P,(\pi_i)_{i\in I})$ of the family exists in $\mathcal{C}$ and let $(\prod_{i\in I}M_i,(p_i)_{i\in I})$ be their cartesian product with the corresponding projections, i.e. their product in $\mathbf{Set}$. By the universl property, there exists a map $f\colon P\rightarrow\prod_{i\in I}M_i$, such that $p_i\circ f=\pi_i$ for all $i\in I$. Let $x,y\in P$ such that $f(x)=f(y)$. Let $\overline{x},\overline{y}\colon\{\ast\}\rightarrow P$ be the morphisms in $\mathcal{C}$ mapping $\ast$ to $x,y$ respectively. Then $f\circ\overline{x}=f\circ\overline{y}$, hence $\pi_i\circ\overline{x}=p_i\circ f\circ\overline{x}=p_i\circ f\circ\overline{y}=\pi_i\circ\overline{y}$ for all $i\in I$, hence $\overline{x}=\overline{y}$ and thus $x=y$, i.e. $f$ is injective. Let $(x_i)_{i\in I}\in\prod_{i\in I}M_i$. For each $i\in I$, let $\overline{x_i}\colon\{\ast\}\rightarrow M_i$ be the morphism in $\mathcal{C}$ mapping $\ast$ to $x_i$. By the universal property, this factors as a map $\varphi\colon\{\ast\}\rightarrow P$, which satisfies $p_i\circ f\circ\varphi=\pi_i\circ\varphi=\overline{x_i}$ for all $i\in I$, i.e. $f(\varphi(\ast))=(x_i)_{i\in I}$, thus $f$ is surjective. So, by transport of structure, we may WLOG assume that $P=\prod_{i\in I}M_i$ (as sets).

The continuity of the maps $\pi_i\colon\prod_{i\in I}M_i\rightarrow M_i$ immediately forces the topology on $P$ to be at least as fine as the product topology. The hypotheses imply that infinitely many ofthe $M_i,i\in I$ are discrete and non-trivial, hence the product topology on $\prod_{i\in I}M_i$ has uncountably many connected components. But the number of connected components behaves monotonically with respect to fineness of topologies, so that $P$ also has uncountably many connected components, contradicting secound-countability.


What you wrote seems correct.

Your second question is then about products where you have infinitely many discrete manifolds, and only a finite number of positive dimension manifolds.

Since products are associative, and you know the product of a finite number of manifolds exists, you are reduced to the product of discrete manifolds.

Of course, since the point is a manifold, if you have such a product, then its underlying set will be the usual cartesian product of the sets.

Therefore if infinitely many of them aren't singletons, then your product is necessarily uncountable. I will prove that it's discrete, which will be a contradiction with second countability.

Indeed, let $M$ be a manifold. Then if $M$ has positive dimension, it receives a non constant smooth/continuous map $\mathbb R\to M$. But of course, in the case of a product of discrete manifolds, any map $\mathbb R\to \prod_i X_i$ must be constant, since its projection on any $X_i$ is constant.

It follows that $\prod_i X_i$ must have dimension $0$ and be uncountable: absurd.

EDIT : Here's how to conclude for the general case where you have infinitely many discrete manifolds with more than one point and finitely many positive dimensional manifolds; let's call the positive dimensional ones $M_1,...,M_n$ and $(X_i)$ the other ones. Let's call $P$ a product of these. Recall that the underlying set of $P$ is the usual product $M_1\times ... \times M_n \times \prod_i X_i$. Now fix $m\in M_1\times...\times M_n$, and consider $(x_i)\in \prod_i X_i$ (here I'm talking about the cartesian products of sets)

Then, any map $\mathbb R\to P$ which hits $(m,(x_i))$ must have its $X_i$-coordinates $x_i$, because the projection $P\to X_i$ is continuous, and the latter is discrete. In other words, if $(x_i)\neq (y_i)$, $(m,(x_i))$ and $(m,(y_i))$ are in different connected components of $P$. In particular, $P$ has at least $|\prod_i X_i|$ connected components, which is too much for a manifold.

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  • $\begingroup$ Sorry for the belated question, but can you elaborate on why it's only necessary to consider products of discrete manifolds? It's not true in full generality that the existence of the product of a family and the existence of the product of a subfamily implies the existence of the product of the complementary subfamily and I'm unsure why it holds here. $\endgroup$ – Thorgott Oct 23 at 12:22
  • $\begingroup$ Well you are correct, and it is true that you want to know which products exactly exist. In that case I suggest you adapt the argument I gave : the story won't be that there are no nonconstant maps from $\mathbb R$, but that there are too few. In other words, you can prove in exactly the same way that you have an uncountable number of connected components, which again is impossible if your manifolds are second countable $\endgroup$ – Maxime Ramzi Oct 23 at 12:24
  • $\begingroup$ I believe I found an argument for the remaining case, which I've appended to the question, but I wasn't able to figure out the proof you had in mind. How can we use maps from $\mathbb{R}$ to determine the number of connected components? $\endgroup$ – Thorgott Oct 29 at 18:29
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    $\begingroup$ I've added a few words in an edit $\endgroup$ – Maxime Ramzi Oct 29 at 19:41

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