Any smooth manifold defines a Frölicher space with curves $C^\infty(\mathbb{R}, M)$ and functions $C^\infty(M, \mathbb{R})$.

Taking quotients in the category of Frölicher spaces is straightforward: the smooth functions are those that pull-back to smooth functions on the original space.

As an example, consider the plane $\mathbb{R}^2$ quotiented out by the $x$-axis.

Let us write this as $X$. This example is closely related to taking cones and suspensions in algebraic topology. The smooth functions on $X$ are simple to describe: the set is equivalent to those smooth functions $\mathbb{R}^2 \to \mathbb{R}$ which are constant on the $x$-axis.

Now let us consider the smooth curves. Let $\alpha : \mathbb{R} \to X$ be a smooth curve. We can partition $\mathbb{R}$ into two pieces: those points that are mapped to the squashed point in $X$ and those points that aren’t. Let us write $A$ for the set of points that are **not** mapped to the squashed point. Using bump functions it is easy to show that $A$ is open in $\mathbb{R}$. As the quotient map $\mathbb{R}^2 \to X$ is bijective off the $x$-axis, the restriction of $\alpha$ to $A$ has a unique lift to $\mathbb{R}^2$. Let us write $\alpha_x$ and $\alpha_y$ for the coordinate functions of this lift. Again using bump functions, it is easy to show that $\alpha_x$ and $\alpha_y$ are smooth on $A$. Furthermore, as the projection $(x,y) \mapsto y$ descends to a smooth function on $X$, $\alpha_y$ is actually the restriction to $A$ of a smooth function $\mathbb{R} \to \mathbb{R}$ which we shall also denote by $\alpha_y$. Note that $A = \alpha_y^{-1}(0)$.

The interesting part comes when looking at what happens to $\alpha_x$ at the boundary of $A$. As $A$ is open, it is a disjoint union of open intervals. The boundaries of each of these intervals forms part of the boundary of $A$ and it is simplest to start with these points. For further simplicity, let us assume that $(0,1)$ is one of the components of $A$ and we are considering the boundary point $0$. Thus we wish to consider $\lim_{t \to 0^+} \alpha_x(t)$.

The general rule is simple to state: $\alpha_y$ must go to zero faster than $\alpha_x$ (and any of its derivatives) can go to infinity.

Let us give an example that shows that the category of Frölicher spaces is not locally cartesian closed. Consider a coequaliser diagram $\mathbb{R} \setminus \{0\} \to \mathbb{R} \amalg \mathbb{R}$ where the two maps are the inclusions into the two cofactors.

The coequaliser of this diagram is $\mathbb{R} \cup \{*\}$ where the $*$ is a doubled-point at $0$. (So this is the well-known example of a non-Hausdorff manifold.) Thus any smooth function $\psi : \mathbb{R} \cup \{*\} \to \mathbb{R}$ has to satisfy $\psi(0) = \psi(*)$, which means that any smooth curve $\alpha : \mathbb{R} \to \mathbb{R} \cup \{*\}$ can choose whether to pass through $0$ or $*$ completely arbitrarily.

We consider this as a coequaliser of spaces over $\mathbb{R}$ by taking the obvious map to $\mathbb{R}$ in each case. The colimit is the same whether we work in the full category of Frölicher spaces or just those over $\mathbb{R}$.

Now let $Y$ be any Frölicher space and consider it as a space over $\mathbb{R}$ via the constant zero map, $\omicron : Y \to \mathbb{R}$. We take the fibred product over $\mathbb{R}$ of the coequaliser diagram. Since $\mathbb{R} \setminus \{0\}$ has no points mapping to $0$, $(\mathbb{R} \setminus \{0\}) \times_{\mathbb{R}} Y = \emptyset$. For the third space, we see that $(\mathbb{R} \amalg \mathbb{R}) \times_{\mathbb{R}} Y = Y \amalg Y$. Thus the coequaliser diagram is now $\emptyset \to Y \amalg Y$. The coequaliser is thus $Y \amalg Y$. Note that smooth curves into $Y \amalg Y$ are of the form $(i, \beta)$ where $i$ is a constant in $\{0,*\}$ that distinguishes the cofactors and $\beta : \mathbb{R} \to Y$ is a smooth curve in $Y$.

Let us consider the product over $\mathbb{R}$ of $\mathbb{R} \cup \{*\}$ with $Y$. As a set, this is just $Y \amalg Y$ again. However, as a Frölicher space it has different functions to those on $Y \amalg Y$. The product over $\mathbb{R}$ is a subspace of $(\mathbb{R} \cup \{*\}) \times Y$ and thus a curve into it is smooth if and only if it is smooth into $(\mathbb{R} \cup \{*\}) \times Y$, whence it is smooth if and only if the projections to $\mathbb{R} \cup \{*\}$ and to $Y$ are smooth. As we are considering curves in $(\mathbb{R} \cup \{*\}) \times_\mathbb{R} Y$, the projection to $\mathbb{R} \cup \{*\}$ must have image in $\{0,*\}$. Thus *any* curve is allowed by this and so the smooth curves into $(\mathbb{R} \cup \{*\}) \times_\mathbb{R} Y$ are of the form $(\alpha, \beta)$ where $\alpha$ is *any* function from $\mathbb{R}$ into $\{0,*\}$ and $\beta : \mathbb{R} \to Y$ is a smooth curve in $Y$.

I notice that in some classically false versions of constructive mathematics, the only functions from $\mathbb{R}$ to $\{0,*\}$ are the constant ones. It would be nice if there were a nonclassical dream universe in which the category of Frölicher spaces were locally cartesian closed! Unfortunately, the counterexample can be saved by using a continuously parametrised coproduct $\coprod_{\mathbb{R}} \mathbb{R}$ instead of $\coprod_{\{0,*\}} \mathbb{R} = \mathbb{R} \amalg \mathbb{R}$. —Toby

Andrew Stacey I think I’d be disappointed if locally cartesian became a property of what set theory was being used! I suspect that the real reason this example works is the fact that $Y \to \mathbb{R}$ has such bad path-lifting properties and the coequaliser diagram that I chose is just a simple one that demonstrates it.

I’ve been pondering how one might fix this. I ought to write up Kriegl and Michor’s extension of a manifold as, although they haven’t done much with it, it contains some ideas that may be of interest. What makes me think of it here is that that definition of a manifold (which isn’t, by the way, the one in their weighty tome) has two parts: a space and its tangent space, and relationships between them. This suggests that a smooth space, of whatever variety, should be more than just one space but some sort of diagram of spaces.

This is also suggested by trying to generalise the Frolicher “idea” to non-set-based theories. It’s not immediately obvious how to make the saturation condition work, but the basic idea is to have objects as $(C,F,c)$ with $c : C \times F \to Hom$ and morphisms as $f : C_X \times F_Y \to Hom$. The saturation condition, whatever it is, is what is needed to make this into a category. So here it’s obvious that objects are just special morphisms. Feeding this back into the definition you get a glimmer of an idea that maybe a morphism (let me go back to genuine Frolicher spaces here for clarity) $f : X \to Y$ needs some sort of saturation condition as well as a compatibility condition.

This is not the same as $Y \coprod Y$ and thus the functor $- \times_{\mathbb{R}} Y$ does not preserve colimits. It cannot, therefore, be a left adjoint and so the category of Frölicher spaces is not locally cartesian closed.

This example works because of the structure of $\mathbb{R} \cup \{*\}$. If one were to work in an “input only” category, then the structure on $\mathbb{R} \cup \{*\}$ would be determined by those curves which lift to $\mathbb{R} \coprod \mathbb{R}$. Such maps could not arbitrarily swap between $0$ and $*$ because up in $\mathbb{R} \coprod \mathbb{R}$ these two points are far apart. Thus the subspace structure on $\{0,*\}$ in an “input only” category is *discrete*. However, in the category of Frölicher spaces the outputs control the behaviour of quotients. Functions out of $\mathbb{R} \cup \{*\}$ cannot detect the difference between $0$ and $*$. Thus curves into $\mathbb{R} \cup \{*\}$ are allowed to swap between them with aplomb. The subspace structure on $\{0,*\}$ is thus the *indiscrete* structure.

Working in the category of *Hausdorff* Frölicher spaces (see below) does not improve matters. Then we need to replace each coequaliser but its Hausdorffification. Now the distinction is clear since taking the product and then the coequaliser yields $Y \coprod Y$ as before but taking the coequaliser and then the product yields just $Y$.

It is also worth pointing out that with the modification of the previous paragraph, this example only involves manifolds (assuming that $Y$ is chosen to be a manifold). It therefore shows that a category extending that of smooth manifolds can **either** be locally cartesian closed **or** preserve limits and colimits from manifolds *but not both*.

Last revised on October 26, 2010 at 05:04:30. See the history of this page for a list of all contributions to it.