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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?

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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

2 Answers 2

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 ($\emptyset$).

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Ok. Good point. –  Mark.Neuhaus Apr 22 '12 at 19:53

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 $f= h\circ \pi_2$.

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