Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $V$ be the category of finite dimensional vector spaces and $M$ the category of smooth finite dimensional Hausdorff manifolds.

Now suppose any finite dimensional vector space is equipped with a smooth structure in such a way that any $n$-dimensional vector space is diffeomorph to $\mathbb{R}^n$ seen as a smooth manifold with the standard smooth structure.

This way there is a faithfull inclusion $\imath: V \to M$ by just forgetting the linear structure.

Now recall that $V$ is cocomplete while $M$ is not.

To see that colimits exist in $V$ let $D : I \to V$ be a diagram with a finite index category $I$. To construct the colimit, let $h_i : D_i \to \bigoplus_{j \in I} D_j$ be the inclusions and $Q$ be the submodule generated by the images of the maps $h_i \circ Dd - h_j$ for each morphism $d : j \to i$, and let $C = \bigoplus_{j\in I} D_j /Q$ be the quotient space. Then $(D_i \overset{q\circ h_i}{\to} C)_{i \in I}$ is a colimit of $D$, where $q$ is the quotient map.

Counterexamples to the existence of all colimits in $M$ are given here on MO for example at: Colimits in the category of smooth manifolds

Now the question is: Does the inclusion $i: V \to M$ preserves these (finite) colimits?

Obviously $(D_i \overset{q\circ h_i}{\to} C)_{i \in I}$ is a cocone in $M$, but is it sill universal?

share|improve this question
    
WHat is the smooth structure on $\oplus_i D_i$ and on $C$ ? I think that this is the true problem inside your question. –  Buschi Sergio Apr 22 '12 at 19:13
    
No. Let each $D_i$ be $n_i$-dimensional. Then $\oplus_iD_i$ is $n:= \sum_i n_i$-dimensional and hence diffeomorph to $\mathbb{R}^n$. Similar for $C$. It is again just a finite dimensional vector space an hence has the apropriate standard smooth structure. Recall that for $n \neq 4$ there simply is just one smooth structure. –  Mark.Neuhaus Apr 22 '12 at 19:21
    
I think your question only asks for finite colimits. In fact, $V$ is only finitely cocomplete. –  Martin Brandenburg Apr 22 '12 at 19:31
    
Any finite dimensional vector space carries a canonical smooth structure in the following manner: If $dim(V)= n$, we take the atlas consisting of all linear isomorphisms $\phi : V \to \mathbb{R}^n$.This collection of maps is an atlas since for any two $\phi$ and $\psi$ the change of coordinates is a linear map $\mathbb{R}^n \to \mathbb{R}^n$ and hence smooth. If $dim(V)=4$ we have in addition to require that we consider the standard smooth structure on $\mathbb{R}^4$ since there is a continuum of others. –  Mark.Neuhaus Apr 22 '12 at 19:33
    
Sure, but then you should write "Does the inclusion $i: V \to M$ preserves finite colimits?" above. –  Martin Brandenburg Apr 22 '12 at 19:35
show 1 more comment

2 Answers

up vote 1 down vote accepted

The canonical map $i(\mathbb{R}^n) \coprod_M i(\mathbb{R}^m) \to i(\mathbb{R}^n \coprod_V \mathbb{R}^m)$, where the coproduct index indicates the ambient category, corresponds to the smooth map $\mathbb{R}^n \sqcup \mathbb{R}^m \to \mathbb{R}^{n+m}$. It is neither surjective nor injective (the two zero vectors are mapped to the zero vector). So $i$ doesn't preserve coproducts.

The problem is already that $i$ maps the initial vector space to the point, which is the terminal manifold, but not the initial manifold ($\emptyset$).

share|improve this answer
    
Ok. Good point. –  Mark.Neuhaus Apr 22 '12 at 19:53
add comment

Mark. You write "Obviously $(D_i\overset{q\circ h_i}{\to}C)_{i\in I}$ is a cocone in $M$, then I ask what you think for smooth structure of these objects, of course about finite sum (that is a biproduts then a product) the answere is as you said "the product of smooth structures", but for $C$?.

I put this countrexample (primarily I consider no the smooth manifolds, but topological spaces):

let $d_1, 0: \mathbb{R}\to \mathbb{R}^2$ the inclusion map $x \mapsto (x, 0)$ (the $X$ axis is the image) and the $0$-costant map. The cokernel in $V$ of these maps is $\mathbb{R}$ (the cartesian axis $Y$), with the projection $\pi_2: \mathbb{R}^2\to \mathbb{R}: (x, y)\mapsto y$. But the cokernel of $i(0), i(\Delta)$ in $Top$ (topological spaces) is like a cone without a line (think the plane as a square without boundary, and by the middle horizontal line as the image of $d_1$, then make with this a (finite) cylinder without a line, then contracting the middle line to a point).

I guess that this cokernel dont exist in the smooth manifolds. Anyway the cokern of $i(0), i(d_1)$ cannot be the projection (to the line $Y$) $\pi_2: \mathbb{R}^2\to i(\mathbb{R})$:

Let $S\subset \mathbb{R}^3$ the rotations parabolid with equation $z= x^2+y^2$ and $f: \mathbb{R}^2\to S: (x, y) \mapsto (x^2\cdot y^2, y^2, y^2(1+x^2) $ (this is smooth, surjective, with injective restriction to the open cartesian quadrants, and send all $X$ axis on $(0,0)$). For topological dimention topics about smooth maps, cannot exist a (surjective) smooth map $h: \mathbb{R}\to S$ with $f= h\circ \pi_2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.