nLab
topological notions of Frölicher spaces

Idea

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 PP 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 PP defines a topological space with the corresponding property PP. 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.

Functors

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

Definition

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.

Lemma

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

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

Proof

We merely need to observe that the family {U:c 1(U)opencC X}\{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)f^{-1}(a,b). That is, these sets form a subbasis? for the topology. Using the structure of the functionals, we can strengthen that.

Lemma

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

Proof

The “if” is obvious so we prove the “only if”. Thus let UXU \subseteq X be open and pUp \in U. Then there are f 1,,f nf_1, \dots, f_n and a i,b ia_i, b_i \in \mathbb{R} such that

p if i 1(a i,b i)U. p \in \bigcap_i f_i^{-1}(a_i,b_i) \subseteq U\;.

We combine the functions f if_i into a single function f(f 1,,f n):X nf \coloneqq (f_1, \dots, f_n) \colon X \to \mathbb{R}^n. Then

if i 1(a i,b i)=f 1(a 1,b 1)××(a n,b n) \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 n\mathbb{R}^n is smoothly regular to find a smooth function g: ng \colon \mathbb{R}^n \to \mathbb{R} such that g(f(p))=1g(f(p)) = 1 and g(y)=0g(y) = 0 outside the specified rectangle. Then the composition gfg \circ f has the required properties.

Basic Properties

Let us start with some very simple definitions.

Definition

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.

Lemma

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.

Proof

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

(1)α(t)={x t0 y t0 \alpha(t) = \begin{cases} x & t \leq 0 \\ y & t \ge 0 \end{cases}

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

(2)t{ϕ(x) t0 ϕ(y) t0 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 (,)C^\infty(\mathbb{R}, \mathbb{R}) we must have ϕ(x)=ϕ(y)\phi(x) = \phi(y), hence ϕ\phi is constant.

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

The discrete case is similar.

Separation Axioms

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.

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.

Lemma

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

Proof

Suppose that (X,C,F)(X,C,F) is not functionally Hausdorff. Then there are xyXx \ne y \in X such that ϕ(x)=ϕ(y)\phi(x) = \phi(y) for all ϕX\phi \in X. Let α:X\alpha : \mathbb{R} \to X be the function taking the value xx for t0t \leq 0 and yy for t0t \ge 0. Then ϕα\phi \circ \alpha is constant for all ϕF\phi \in F so αC\alpha \in C. However, α\alpha has finite image but is not constant. Thus (X,C,F)(X,C,F) is not curvaceously Hausdorff.

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

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

Suppose that XX with the curvaceous topology is Hausdorff. Then any finite subset is discrete and so there are no non-constant continuous maps X\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.

Lemma

Let (X,C,F)(X,C,F) be a Frölicher space. Let YY be the quotient of XX by the relation xyx \sim y if ϕ(x)=ϕ(y)\phi(x) = \phi(y) for all ϕF\phi \in F. Then YY inherits a Frölicher space structure from XX with respect to which it is Hausdorff. The natural map XYX \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 XX.

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

Proof

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

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

That this is a quotient is straightforward. Any smooth map g:XZg : X \to Z which factors through YY as a set must also do so as a Frölicher space. In particular, if g:XZg : X \to Z is a smooth map with ZZ Hausdorff then this must factor through YY 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 YY choose a representative of the equivalence class. This choice defines a map on the underlying sets YXY \to X. This is also smooth since the sets of functions F XF_X and F YF_Y are identified by the quotient mapping XYX \to Y.

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

This shows, incidentally, that every smooth curve in the Hausdorffification of XX lifts to a smooth curve in XX. 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.

Lemma

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

Proof

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

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

Definition

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.

Compactness

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

Definition

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.

Lemma

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

Proof

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)(x_n) with no accumulation points. As the functional topology is (smoothly) regular, we can find smooth functions (ϕ n)(\phi_n) such that ϕ n(x m)=δ nm\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 nϕ n^\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.

Lemma

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

Proof

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 ×X\mathbb{R} \times X. For Frölicher spaces, this property is equivalent to sequential compactness.

Proposition

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

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

Proof

Suppose that XX is sequentially compact. Let U×XU \subseteq \mathbb{R} \times X be a subset containing {0}×X\{0\} \times X. Suppose that UU does not contain a subset of the form (ϵ,ϵ)×X(-\epsilon,\epsilon) \times X. Then for each nn \in \mathbb{N} we can find some (t n,x n)U(t_n, x_n) \in U such that abst n1n\abs{t_n} \le \frac1n. The sequence (x n)(x_n) in XX has a convergent subsequence, say (x n k)x(x_{n_k}) \to x. Then (t n k,x n k)(t_{n_k}, x_{n_k}) converges in ×X\mathbb{R} \times X, but (t n k,x n k)U(t_{n_k}, x_{n_k}) \notin U for all kk but its limit, (0,x)(0,x) is in UU. Hence UU is not open.

Conversely, assume that neighbourhoods of {0}×X\{0\} \times X contain slices as claimed. Then let (x n)(x_n) be a sequence in XX. Consider the set UU formed by taking ×X\mathbb{R} \times X and removing (1n,x n)(\frac1n, x_n). This does not contain a subset of the form (ϵ,ϵ)×X(-\epsilon,\epsilon) \times X and so cannot be open in ×X\mathbb{R} \times X.

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

Connectedness

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

Definition

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)(X,C,F) are the equivalence classes of the relation xyx \sim y if whenever ϕF\phi \in F is idempotent then ϕ(x)=ϕ(y)\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)(X,C,F) are the equivalence classes of the relation xyx \sim y if there is a smooth curve αC\alpha \in C with α(0)=x\alpha(0) = x and α(1)=y\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.

Lemma

The notions of functionally connected and curvaceously connected coincide.

Proof

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

Corollary

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

More on Hausdorff Frölicher Spaces

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\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.

Theorem

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 𝔉:I\mathfrak{F} : I \to \mathcal{M} be a functor where II is a small category.

Let us assume first that 𝔉\mathfrak{F} has a limit in \mathcal{M}, say M 0M_0 with maps α i:M 0𝔉(i)\alpha_i : M_0 \to \mathfrak{F}(i). Let us write X 0X_0 for the limit of 𝔉\mathfrak{F} viewed as a functor into \mathcal{H}, with maps β i:X 0𝔉(i)\beta_i : X_0 \to \mathfrak{F}(i). As \mathcal{H} is a reflective subcategory of \mathcal{F}, X 0X_0 is the same as the limit of 𝔉\mathfrak{F} in \mathcal{F}.

Since M 0M_0, as a Frölicher space, is a source of 𝔉\mathfrak{F}, there is a unique map, say γ:M 0X 0\gamma : M_0 \to X_0, such that β iγ=α i\beta_i \gamma = \alpha_i.

As a Frölicher space, X 0X_0 is completely determined by its underlying set and its smooth curves. Its underlying set is (naturally isomorphic to) (*,X 0)\mathcal{F}(*,X_0). Let x|X 0|x \in |X_0|. Composing with the β i\beta_i defines maps β ix:*𝔉\beta_i x : * \to \mathfrak{F}. Since ** is a manifold and M 0M_0 is the limit of 𝔉\mathfrak{F} in \mathcal{M}, there is a unique map x^:*M 0\widehat{x} : * \to M_0 such that α ix^=β ix\alpha_i \widehat{x} = \beta_i x for all iIi \in I. Using the uniqueness of the factorisations, we see that γx^=x\gamma \widehat{x} = x and thus γ\gamma induces a bijection |M 0||X 0||M_0| \to |X_0|. Hence the underlying sets of M 0M_0 and X 0X_0 are the same.

The smooth curves of X 0X_0 are the morphisms X 0\mathbb{R} \to X_0. Since \mathbb{R} is a manifold, the same argument shows that γ\gamma induces a bijection from the set of smooth curves in M 0M_0 to that in X 0X_0. Hence γ\gamma is an isomorphism of Frölicher spaces and so the inclusion functor \mathcal{M} \to \mathcal{H} preserves limits.

Now let us assume that 𝔉\mathfrak{F} has a colimit in \mathcal{M}, say M 1M_1 with maps λ i:𝔉(i)M 1\lambda_i : \mathfrak{F}(i) \to M_1. Let us write X 1X_1 for the colimit of 𝔉\mathfrak{F} viewed as a functor into \mathcal{F}, with maps μ i:mathfralF(i)X 1\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 1X_1.

Since M 1M_1, as a Hausdorff Frölicher space, is a sink of 𝔉\mathfrak{F} there is a unique morphism, say ν:X 1M 1\nu : X_1 \to M_1, such that νμ i=λ i\nu \mu_i = \lambda_i. This morphism factors uniquely through the Hausdorffification of X 1X_1.

For the same argument as with the limits, the smooth functions on X 1X_1 factor through those of M 1M_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 ν\nu induces an isomorphism from the Hausdorffification of X 1X_1 to M 1M_1.

Firstly, let us show that ν\nu is surjective on underlying sets. To see this, suppose for a contradiction that it is not. Let x|M 1|x \in |M_1| be a point not in the image of ν\nu. Let N=M 1{x}N = M_1 \smallsetminus \{x\}. Then NN is an open submanifold of M 1M_1 and ν\nu factors through the inclusion NM 1N \to M_1. As X 1X_1 is the colimit of 𝔉\mathfrak{F}, the morphism X 1NX_1 \to N establishes NN as a sink for 𝔉\mathfrak{F}. As NN is a manifold, there is thus a unique morphism M 1NM_1 \to N factoring the morphisms from 𝔉\mathfrak{F} to NN. That is to say, the morphism X 1NX_1 \to N uniquely factors through ν:X 1M 1\nu : X_1 \to M_1. This gives a factorisation of ν\nu as

X 1 νM 1NM 1 X_1 \longrightarrow^{\nu} M_1 \to N \to M_1

where the last morphism is the inclusion of NN in M 1M_1. However, the properties of ν\nu imply that the morphism M 1M 1M_1 \to M_1 in the above diagram is the identity, contradicting the non-surjectivity of NM 1N \to M_1 and thus the non-surjectivity of X 1M 1X_1 \to M_1.

Hence X 1M 1X_1 \to M_1 is surjective. We also have that the smooth functions on X 1X_1 factor through M 1M_1. This is not enough to prove that X 1X_1 and M 1M_1 are isomorphic, but is enough to prove that the Hausdorffification of X 1X_1 is isomorphic to M 1M_1 (note that M 1M_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 ν\nu induces an injective map from the underlying set of the Hausdorffification of X 1X_1 to the underlying set of M 1M_1. Thus let x,yx, y be distinct points in the Hausdorffification of X 1X_1. There is thus a smooth function on X 1X_1 which distinguishes them. As this smooth function factors through M 1M_1, we must have ν(x)ν(y)\nu(x) \ne \nu(y) and hence ν\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 {0}\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}\{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 XX be the real line with a double point at the origin. Take a curve X\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.

Revised on September 16, 2013 13:04:04 by Raeder? (129.241.15.244)