Although the definition of a Frölicher space does not use a topology, it is topological in flavour and there are many topological concepts that can be defined for Frölicher spaces. In other pages, we have used the notion of a Hausdorff Frölicher space and have used a vague notion of a topology on the functions on a Frölicher space. In this page, we shall investigate the connections between the theory of Frölicher spaces and topological spaces a little more consistently.

There are two different ways of thinking of topological notions on Frölicher spaces. One says that there is a functor (actually two functors) from the category of Frölicher space to the category of topological spaces so we can say that a Frölicher space has topological property $P$ if the corresponding topological space has it. The other approach says that we can directly define a property for Frölicher spaces that is *analogous* to a topological property. We then might hope for a theorem saying that a Frölicher space with property $P$ defines a topological space with the corresponding property $P$. However, this would definitely be a theorem.

The first approach involves defining a functor (or more than one) from the category of Frölicher spaces to that of topological spaces and looking at the image. The second approach involves comparing the two categories and trying to transfer ideas from the more well-known category of topological spaces to that of Frölicher spaces. We shall study both.

Let us start by defining the two functors to topological spaces.

The *curvaceous* topology on a Frölicher space is the strongest topology for which the smooth curves are continuous.

The *functional* topology on a Frölicher space is the weakest topology for which the smooth functions are continuous.

It is clear that these assignments are functorial, and that the curvaceous topology is always at least as strong as the functional topology.

As it is an inductive topology, the curvaceous topology has the following characterisation.

Let $(X,C_X,F_X)$ be a Frölicher space. A subset $U \subseteq X$ is open in the curvaceous topology if and only if $c^{-1}(U)$ is open in $\mathbb{R}$ for all $c \in C_X$.

Thus a subset $A \subseteq X$ is not open if and only if there is some $c \in C_X$ such that $c^{-1}(A)$ is not open in $\mathbb{R}$.

We merely need to observe that the family $\{U : c^{-1}(U) \;\text{open}\; \forall c \in C_X\}$ defines a topology. This is elementary.

The functional topology is a projective topology and so it is *generated* by sets of the form $f^{-1}(a,b)$. That is, these sets form a subbasis? for the topology. Using the structure of the functionals, we can strengthen that.

Let $(X,C_X,F_X)$ be a Frölicher space. A subset $U \subseteq X$ is open in the functional topology if and only if for each $p \in U$ there is a function $f \in F_X$ such that $f(p) = 1$ and $f(q) = 0$ for $q \notin U$.

The “if” is obvious so we prove the “only if”. Thus let $U \subseteq X$ be open and $p \in U$. Then there are $f_1, \dots, f_n$ and $a_i, b_i \in \mathbb{R}$ such that

$p \in \bigcap_i f_i^{-1}(a_i,b_i) \subseteq U\;.$

We combine the functions $f_i$ into a single function $f \coloneqq (f_1, \dots, f_n) \colon X \to \mathbb{R}^n$. Then

$\bigcap_i f_i^{-1}(a_i,b_i) = f^{-1} (a_1,b_1) \times \dots \times (a_n,b_n)$

Now we use the fact that the topology on $\mathbb{R}^n$ is smoothly regular to find a smooth function $g \colon \mathbb{R}^n \to \mathbb{R}$ such that $g(f(p)) = 1$ and $g(y) = 0$ outside the specified rectangle. Then the composition $g \circ f$ has the required properties.

Let us start with some very simple definitions.

A Frölicher space is said to be *indiscrete* if all curves are smooth.

A Frölicher space is said to be *discrete* if all functions are smooth.

Let us observe that there is no need for functional or curvaceous versions of these definitions.

A Frölicher space is indiscrete if and only if the only smooth functions are the constant ones.

A Frölicher space is discrete if and only if the only smooth curves are the constant ones.

If all curves are smooth then for any points $x, y \in X$ the curve

(1)$\alpha(t) = \begin{cases}
x & t \leq 0 \\
y & t \ge 0
\end{cases}$

is smooth. Composition with $\phi \in F$ yields the function

(2)$t \mapsto \begin{cases}
\phi(x) & t \leq 0 \\
\phi(y) & t \ge 0
\end{cases}$

For this to be a smooth function in $C^\infty(\mathbb{R}, \mathbb{R})$ we must have $\phi(x) = \phi(y)$, hence $\phi$ is constant.

For the converse, if the only smooth functions are constant then any curve $\alpha : \mathbb{R} \to X$ satisfies the condition that $\phi \circ \alpha \in C^\infty(\mathbb{R}, \mathbb{R})$ for all $\phi \in F$ so is in $C$. Hence all curves are smooth.

The discrete case is similar.

In other sections, namely examples of Frölicher spaces and Frölicher spaces and Isbell envelopes we used the notion of a *Hausdorff* Frölicher space. Technically, we ought to have called that *functionally Hausdorff* as it used the smooth functions in its definition.

A Frölicher space is said to be *functionally Hausdorff* if the smooth functions separate points.

A Frölicher space is said to be *curvaceously Hausdorff* if the only smooth curves with finite image are constant.

However, the distinction is not important as the following lemma shows.

The notions of functional Hausdorff and curvaceously Hausdorff coincide and are equivalent to the underlying topological spaces being Hausdorff.

Suppose that $(X,C,F)$ is not functionally Hausdorff. Then there are $x \ne y \in X$ such that $\phi(x) = \phi(y)$ for all $\phi \in X$. Let $\alpha : \mathbb{R} \to X$ be the function taking the value $x$ for $t \leq 0$ and $y$ for $t \ge 0$. Then $\phi \circ \alpha$ is constant for all $\phi \in F$ so $\alpha \in C$. However, $\alpha$ has finite image but is not constant. Thus $(X,C,F)$ is not curvaceously Hausdorff.

Conversely, suppose that $(X,C,F)$ is not curvaceously Hausdorff. Then there is some $\alpha \in C$ with finite image which is not constant. Let $\phi \in F$. Then $\phi \circ \alpha$ has finite image in $\mathbb{R}$ and hence is constant. Thus for $x, y \in \im \alpha$, $\phi(x) = \phi(y)$ for all $\phi \in F$. As $\alpha$ is not constant, there are thus $x \ne y \in X$ such that $\phi(x) = \phi(y)$ for all $\phi \in F$ and so $(X,C,F)$ is not functionally Hausdorff.

If a Frölicher space is Hausdorff then smooth functions separate points. Thus for $x \ne y \in X$, there is a smooth function $\phi \in F$ with $\phi(x) = -1$ and $\phi(y) = 1$. Then the sets $\phi^{-1}(-\infty,0)$ and $\phi^{-1}(0,\infty)$ are sufficient to show that $X$ with the functional topology is Hausdorff. As the curvaceous topology is stronger than the functional one, it is thus also Hausdorff.

Suppose that $X$ with the curvaceous topology is Hausdorff. Then any finite subset is discrete and so there are no non-constant continuous maps $\mathbb{R} \to X$ with finite image. In particular, there are no non-constant smooth maps and so the original Frölicher space was Hausdorff.

In light of this, we shall refer to just *Hausdorff* Frölicher spaces.

Just as with topological spaces, there is a “Hausdorffification” functor. Unlike topological spaces, this functor is split.

Let $(X,C,F)$ be a Frölicher space. Let $Y$ be the quotient of $X$ by the relation $x \sim y$ if $\phi(x) = \phi(y)$ for all $\phi \in F$. Then $Y$ inherits a Frölicher space structure from $X$ with respect to which it is Hausdorff. The natural map $X \to Y$ is a quotient mapping in the category of Frölicher spaces. It is split, but not canonically so. However, any two splittings are related by a diffeomorphism on $X$.

The assignment $X \mapsto Y$ is left adjoint to the inclusion of the category of Hausdorff Frölicher spaces in the category of all Frölicher spaces.

The Frölicher structure on $Y$ is defined by setting $F_Y$ to be the set of functions $\phi : Y \to \mathbb{R}$ such that the composition $X \to Y \xrightarrow{\phi} \mathbb{R}$ is in $F_X$. The smooth curves are then defined by the saturation condition. It is automatic from this definition that any smooth curve in $X$ projects down to a smooth curve in $Y$ which explains why this family of functions on $Y$ is also saturated and hence we have a Frölicher space structure on $Y$.

To show that $Y$ is Hausdorff, we merely observe that by slight abuse of notation, $F_X = F_Y$ so if $x,y \in X$ are such that $\phi(\overline{x}) = \phi(\overline{y})$ for all $\phi \in F_Y$ then $\phi(x) = \phi(y)$ for all $\phi \in F_X$, whence $\overline{x} = \overline{y}$ in $Y$.

That this is a quotient is straightforward. Any smooth map $g : X \to Z$ which factors through $Y$ *as a set* must also do so as a Frölicher space. In particular, if $g : X \to Z$ is a smooth map with $Z$ Hausdorff then this must factor through $Y$ as a set, whence as a Frölicher space. This also establishes the necessary adjunction.

Finally, let us look at the splitting. For each point in $Y$ choose a representative of the equivalence class. This choice defines a map on the underlying sets $Y \to X$. This is also smooth since the sets of functions $F_X$ and $F_Y$ are identified by the quotient mapping $X \to Y$.

Given two such splittings, say $i, j : Y \to X$, define a bijection $X \to X$ which interchanges $i(y)$ and $j(y)$. As $i$ and $j$ are splits of the quotient mapping $X \to Y$ this is well-defined. It is also clearly a diffeomorphism since $F_X$ cannot detect the difference between $i$ and $j$.

This shows, incidentally, that every smooth curve in the Hausdorffification of $X$ lifts to a smooth curve in $X$. This sort of behaviour does not *usually* happen with quotients in the category of Frölicher spaces.

The fibres of the Hausdorffification are straightforward to identify.

The fibres of the Hausdorffification of a Frölicher space correspond precisely to the maximal subsets which inherit an indiscrete structure from the ambient space.

Let $X$ be a Frölicher space, $Y$ its Hausdorffification. For $y \in Y$, let $X_y$ be the corresponding fibre. Then $X_y$ inherits a Frölicher space structure from its inclusion in $X$. The smooth curves in $X_y$ are those that are smooth when considered as curves in $X$. Let $\alpha : \mathbb{R} \to X_y$ be an arbitrary curve. Then for $\phi \in F_X$, as $\im \alpha \subseteq X_y$, $\phi \circ \alpha$ is constant. Hence $\alpha$ is smooth as a curve in $X$, and thus in $X_y$. Thus $X_y$ is indiscrete.

Conversely, let $Z \subseteq X$ be a subset that inherits an indiscrete structure from $X$. Let $x,y \in Z$. Then there is a smooth curve $\alpha$ in $Z$ with $\im \alpha = \{x,y\}$. This is then smooth in $X$ so for all $\phi \in F_X$, $\phi(x) = \phi(y)$. Hence $Z$ is contained in a (unique) fibre of the quotient map $X \to Y$.

Thus when we pass to the Hausdorffification we lose almost no information at all and one could certainly say that we lose no *interesting* information.

Having dealt with Hausdorff Frölicher spaces, the obvious next thing to do is to consider the other separation properties. Our next definition may be a little surprising at first.

A Frölicher space is said to be *regular* if the curvaceous and functional topologies agree.

The point of this definition is that for the underlying topological space of a Frölicher space what one really wants to know is not whether or not it is regular but whether or not it is *smoothly* regular. This is automatic for the functional topology so the only reasonable question is whether or not it happens for the curvaceous topology. However, a topology is smoothly regular if and only if the smooth functions generate the topology which means that the curvaceous topology is smoothly regular if and only if it agrees with the functional topology. Hence the definition.

It is straightforward to see what one version of compactness should be.

A Frölicher space is *functionally compact* if every smooth function has bounded image.

The images are automatically compact as a smooth function with non-compact image can be converted to a smooth function with unbounded image by suitable composition.

A Frölicher space is functionally compact if and only if the functional topology is compact.

One way is obvious: if the functional topology is compact then as the smooth functions are continuous, they have compact image, hence bounded.

For the converse, assume that the functional topology is not compact. Then we can find a countable family of points $(x_n)$ with no accumulation points. As the functional topology is (smoothly) regular, we can find smooth functions $(\phi_n)$ such that $\phi_n(x_m) = \delta_{n m}$. We claim that it is possible to modify these to have disjoint support. This is done recursively using postcomposition by suitably chosen functions. Once this is done, we can define a new smooth function by $\sum n \widehat{\phi_n}$. This is smooth, as the components have disjoint support, and is unbounded. Hence the Frölicher space is not functionally compact.

The curvaceous topology of a Frölicher space is compact if and only if the Frölicher space is functionally compact and regular.

Since the topologies on a Frölicher space are the pull-backs of the topologies on the Hausdorffification, it is sufficient to prove this for a Hausdorff Frölicher space.

As the curvaceous topology is stronger than the functional, if the curvaceous topology is compact then so is the functional. Moreover, as both are Hausdorff spaces, the identity map is a continuous bijection from a compact space to a Hausdorff space and hence a homeomorphism. Thus the Frölicher space is regular.

Conversely, if the Frölicher space is regular its topologies agree and thus if it is functionally compact then its curvaceous topology is compact.

In manifolds of mapping spaces, there is an important issue as to compactness of the actual topological space, rather than compactness notions of Frölicher spaces. The property needed is about open sets in the product $\mathbb{R} \times X$. For Frölicher spaces, this property is equivalent to sequential compactness.

Let $X$ be a Frölicher space. The curvaceous topology on $X$ is sequentially compact if and only if $\mathbb{R} \times X$ has the following property.

A subset $U \subseteq \mathbb{R} \times X$ containing $\{0\} \times X$ is a neighbourhood of $\{0\} \times X$ if and only if it contains a subset of the form $(-\epsilon,\epsilon) \times X$ for some $\epsilon \gt 0$.

Suppose that $X$ is sequentially compact. Let $U \subseteq \mathbb{R} \times X$ be a subset containing $\{0\} \times X$. Suppose that $U$ does not contain a subset of the form $(-\epsilon,\epsilon) \times X$. Then for each $n \in \mathbb{N}$ we can find some $(t_n, x_n) \in U$ such that $\abs{t_n} \le \frac1n$. The sequence $(x_n)$ in $X$ has a convergent subsequence, say $(x_{n_k}) \to x$. Then $(t_{n_k}, x_{n_k})$ converges in $\mathbb{R} \times X$, but $(t_{n_k}, x_{n_k}) \notin U$ for all $k$ but its limit, $(0,x)$ is in $U$. Hence $U$ is not open.

Conversely, assume that neighbourhoods of $\{0\} \times X$ contain slices as claimed. Then let $(x_n)$ be a sequence in $X$. Consider the set $U$ formed by taking $\mathbb{R} \times X$ and removing $(\frac1n, x_n)$. This does not contain a subset of the form $(-\epsilon,\epsilon) \times X$ and so cannot be open in $\mathbb{R} \times X$.

As the set $U$ is not open, there must be a curve $c \colon \mathbb{R} \to U$ which detects the fact that it is not open. That is, $c^{-1}(U)$ is not open in $\mathbb{R}$. Now we can find a sequence $(s_k) \to s$ in $\mathbb{R}$ such that $(s_k) \notin c^{-1}(U)$ but $s \in U$. Then as $c(s_k) \notin U$, we must have $c(s_k) = (t_{n_k}, x_{n_k})$ for some $n_k$. By passing to a subsequence if necessary, we can assume that the $n_k$s are strictly increasing. Then as $(s_k) \to s$, $c(s_k) \to c(s)$ and so $x_{n_k} \to p_X c(s)$. Hence $X$ is sequentially compact.

Another obvious topological property is connectedness. Here it is obvious what the two definitions should be.

A Frölicher space is *functionally connected* if the only idempotents in its algebra of functions are the trivial ones.

More generally, the *functional connected components* of a Frölicher space $(X,C,F)$ are the equivalence classes of the relation $x \sim y$ if whenever $\phi \in F$ is idempotent then $\phi(x) = \phi(y)$.

A Frölicher space is *curvaceously connected* if every pair of points lie on a curve.

More generally, the *curvaceous connected components* of a Frölicher space $(X,C,F)$ are the equivalence classes of the relation $x \sim y$ if there is a smooth curve $\alpha \in C$ with $\alpha(0) = x$ and $\alpha(1) = y$.

That the second relation is an equivalence relation follows from the fact that piecewise smooth curves can be reparametrised to smooth curves.

The notions of functionally connected and curvaceously connected coincide.

Let $(X,C,F)$ be a Frölicher space. It is clear that if $x, y \in X$ are such that there is a smooth curve connecting them then $\phi(x) = \phi(y)$ for any idempotent $\phi \in F$. Thus we need to show the reverse implication. To do this, let $X' \subseteq X$ be a curvaceously connected component of $X$. Let $\phi \colon X \to \mathbb{R}$ be the characteristic function of $X'$. Then for any $\alpha \in C$, either $\im \alpha \subseteq X'$ or $\im \alpha \cap X' = \emptyset$. Thus $\phi \circ \alpha$ is a constant function. Hence $\phi \in F$. Thus if $x$ and $y$ are in different curvaceously connected components there is an idempotent element of $F$ separating them. Hence the two notions are the same.

The functional and curvaceous topologies have the same connected components, and these are the same as the path-connected components.

There is not a great deal of difference between a Hausdorff Frölicher space and a generic one. Much less than the case with topological spaces. To pass from all Frölicher spaces to Hausdorff Frölicher spaces involves only collapsing everything that is *indiscrete*. This clears out a considerable amount of junk from the category but does remove two properties: it no longer has a weak subobject classifier and it is no longer topological over $\Set$.

On the other hand, the relationship between the category of Hausdorff Frölicher spaces and that of all Frölicher spaces is very good. Not only is it a reflective subcategory with all that that implies, but the morphisms from the unit natural transformation are split epimorphisms (though not naturally split).

The category of Hausdorff Frölicher spaces is thus complete and co-complete. It is also cartesian closed since the product and exponential objects of Hausdorff Frölicher spaces are again Hausdorff.

Do I need to prove this, or is it automatic? (I can prove it if necessary)

Mike: Completeness and cocompleteness are of course automatic. It is not automatic that a reflective subcategory inherits cartesian closure; the closest thing I can think of is A4.3.1 in the Elephant which says that a reflective subcategory is an exponential ideal iff its reflector preserves finite products.

Andrew: Is it true, then, that all I need to do is to prove that the exponential of one Hausdorff object by another Hausdorff object is again Hausdorff?

Mike: Yes, that would certainly suffice to show that the category of Hausdorff objects is cartesian closed.

Andrew: Ah, and now I see from exponential ideal that I could just show that the Hausdorffification of a product is the product of the Hausdorffifications. Both seem quite simple, not sure which is the simplest.

In considering Isbell duality in the context of Frölicher spaces we saw one good reason to restrict to Hausdorff Frölicher spaces. Another reason comes from the inclusion of the category of manifolds in that of Frölicher spaces. This factors through Hausdorff Frölicher spaces and this inclusion has some very pleasant properties.

The inclusion functor from the category of Manifolds to that of Hausdorff Frölicher spaces preserves limits and colimits.

Let us write $\mathcal{M}$ for the category of manifolds, $\mathcal{H}$ for the category of Hausdorff Frölicher spaces, and $\mathcal{F}$ for the category of all Frölicher spaces. We shall not give the inclusion functors special symbols but trust to context to distinguish. Let $\mathfrak{F} : I \to \mathcal{M}$ be a functor where $I$ is a small category.

Let us assume first that $\mathfrak{F}$ has a limit in $\mathcal{M}$, say $M_0$ with maps $\alpha_i : M_0 \to \mathfrak{F}(i)$. Let us write $X_0$ for the limit of $\mathfrak{F}$ viewed as a functor into $\mathcal{H}$, with maps $\beta_i : X_0 \to \mathfrak{F}(i)$. As $\mathcal{H}$ is a reflective subcategory of $\mathcal{F}$, $X_0$ is the same as the limit of $\mathfrak{F}$ in $\mathcal{F}$.

Since $M_0$, as a Frölicher space, is a source of $\mathfrak{F}$, there is a unique map, say $\gamma : M_0 \to X_0$, such that $\beta_i \gamma = \alpha_i$.

As a Frölicher space, $X_0$ is completely determined by its underlying set and its smooth curves. Its underlying set is (naturally isomorphic to) $\mathcal{F}(*,X_0)$. Let $x \in |X_0|$. Composing with the $\beta_i$ defines maps $\beta_i x : * \to \mathfrak{F}$. Since $*$ is a manifold and $M_0$ is the limit of $\mathfrak{F}$ in $\mathcal{M}$, there is a unique map $\widehat{x} : * \to M_0$ such that $\alpha_i \widehat{x} = \beta_i x$ for all $i \in I$. Using the uniqueness of the factorisations, we see that $\gamma \widehat{x} = x$ and thus $\gamma$ induces a bijection $|M_0| \to |X_0|$. Hence the underlying sets of $M_0$ and $X_0$ are the same.

The smooth curves of $X_0$ are the morphisms $\mathbb{R} \to X_0$. Since $\mathbb{R}$ is a manifold, the same argument shows that $\gamma$ induces a bijection from the set of smooth curves in $M_0$ to that in $X_0$. Hence $\gamma$ is an isomorphism of Frölicher spaces and so the inclusion functor $\mathcal{M} \to \mathcal{H}$ preserves limits.

Now let us assume that $\mathfrak{F}$ has a colimit in $\mathcal{M}$, say $M_1$ with maps $\lambda_i : \mathfrak{F}(i) \to M_1$. Let us write $X_1$ for the colimit of $\mathfrak{F}$ viewed as a functor into $\mathcal{F}$, with maps $\mu_i \colon \mathfral{F}(i) \to X_1$. Note that this is in $\mathcal{F}$ not $\mathcal{H}$. To obtain the colimit in $\mathcal{H}$ we apply the reflector functor (Hausdorffification) to $X_1$.

Since $M_1$, as a Hausdorff Frölicher space, is a sink of $\mathfrak{F}$ there is a unique morphism, say $\nu : X_1 \to M_1$, such that $\nu \mu_i = \lambda_i$. This morphism factors uniquely through the Hausdorffification of $X_1$.

For the same argument as with the limits, the smooth functions on $X_1$ factor through those of $M_1$. However, the underlying set functor is not represented by morphisms *into* a smooth manifold so we have to be a little more careful to see that the $\nu$ induces an isomorphism from the Hausdorffification of $X_1$ to $M_1$.

Firstly, let us show that $\nu$ is surjective on underlying sets. To see this, suppose for a contradiction that it is not. Let $x \in |M_1|$ be a point not in the image of $\nu$. Let $N = M_1 \smallsetminus \{x\}$. Then $N$ is an open submanifold of $M_1$ and $\nu$ factors through the inclusion $N \to M_1$. As $X_1$ is the colimit of $\mathfrak{F}$, the morphism $X_1 \to N$ establishes $N$ as a sink for $\mathfrak{F}$. As $N$ is a manifold, there is thus a unique morphism $M_1 \to N$ factoring the morphisms from $\mathfrak{F}$ to $N$. That is to say, the morphism $X_1 \to N$ uniquely factors through $\nu : X_1 \to M_1$. This gives a factorisation of $\nu$ as

$X_1 \longrightarrow^{\nu} M_1 \to N \to M_1$

where the last morphism is the inclusion of $N$ in $M_1$. However, the properties of $\nu$ imply that the morphism $M_1 \to M_1$ in the above diagram is the identity, contradicting the non-surjectivity of $N \to M_1$ and thus the non-surjectivity of $X_1 \to M_1$.

Hence $X_1 \to M_1$ is surjective. We also have that the smooth functions on $X_1$ factor through $M_1$. This is not enough to prove that $X_1$ and $M_1$ are isomorphic, but is enough to prove that the Hausdorffification of $X_1$ is isomorphic to $M_1$ (note that $M_1$, being a manifold, is already Hausdorff as a Frölicher space). To see this, observe that with what we already have, all that remains is to show that $\nu$ induces an injective map from the underlying set of the Hausdorffification of $X_1$ to the underlying set of $M_1$. Thus let $x, y$ be distinct points in the Hausdorffification of $X_1$. There is thus a smooth function on $X_1$ which distinguishes them. As this smooth function factors through $M_1$, we must have $\nu(x) \ne \nu(y)$ and hence $\nu$ is injective on the required underlying sets.

Thus the inclusion functor $\mathcal{M} \to \mathcal{H}$ preserves colimits.

The inclusion of the category of Manifolds in that of all Frölicher spaces preserves limits (by the same proof as above) but not colimits. However, it is thus only the issue of being Hausdorff that prevents it preserving colimits. The simplest example is the classic non-Hausdorff manifold: consider the coequaliser of $\mathbb{R} \smallsetminus \{0\}$ included in each piece of $\mathbb{R} \coprod \mathbb{R}$. The colimit in the category of Manifolds is simply $\mathbb{R}$. The colimit of this in the category of Frölicher spaces is the real line with a double point at $\{0\}$, but upon Hausdorffification this becomes the real line.

Mike: What if we re-define “manifold” to remove the Hausdorff axiom? Does the inclusion into Frölicher spaces then preserve colimits?

Andrew: I think so, but the inclusion from manifolds to Frölicher spaces is then not full. Let $X$ be the real line with a double point at the origin. Take a curve $\mathbb{R} \to X$ which oscillates between the two points. This is a morphism into the Frölicher space, but not into the manifold.

I think that to make it work, you have to redefine “Euclidean space” to include anything that becomes a Euclidean space upon Hausdorffification.

Last revised on September 16, 2013 at 13:04:04. See the history of this page for a list of all contributions to it.