1
$\begingroup$

It is easily shown that the forgetful functor $F: \mathbf{Man} \to \mathbf{Set}$ preserves limits ($F$ is representable), but does it preserve colimits? It certainly preserves all examples of colimits that know of which exist in $\mathbf{Man}$, namely

  • coproducts (disjoint unions)
  • the glueing of open manifolds along open subsets
  • quotients obtained by acting properly and freely by Lie groups on manifolds.

The second two cases are special cases of Theorem 5.9.5 in Bourbaki, Variétés differentielles et analytiques:
The set theoretic quotient of a manifold $X$ by an equivalence relation $R \subseteq X \times X$ admits a (necessarily unique) smooth structure such that $X \twoheadrightarrow X/R$ is submersion iff $R \subseteq X \times X$ is a submanifold of $X$ and either projection $R \twoheadrightarrow X$ is a submersion.

By the fact that $F$ preserves coproducts it is enough to show that it preserves coequalisers, but I'm completely stuck on this special case.

$\endgroup$
1
  • $\begingroup$ Both Zhen Lin's and my answer involve non-connected manifolds. It would interesting to know whether there is an example where both manifolds in the coequaliser are connected. $\endgroup$ Commented Feb 3, 2016 at 21:36

2 Answers 2

2
$\begingroup$

In other words, you are asking if there are "exotic" coequalisers in the category of manifolds. In fact, there are.

Let $S^1 = \{ z \in \mathbb{C} : \left| z \right| = 1 \}$ be the circle, let $R = \mathbb{Z} \times S^1$, and let $d_0, d_1 : R \to S^1$ be defined as follows: $$d_0 (n, z) = \exp (i n) z$$ $$d_1 (n, z) = z$$ It is not hard to see that the image of $R \to S^1 \times S^1$ is dense, so if $M$ is a manifold and $f : S^1 \to M$ is a smooth (or even continuous) map such that $f \circ d_0 = f \circ d_1$, then $f : S^1 \to M$ must be a constant map. Hence, the coequaliser of $d_0, d_1 : R \to S^1$ in the category of manifolds is the point. On the other hand, the coequaliser of $d_0, d_1 : R \to S^1$ in the category of sets is an uncountable set. So this coequaliser is not preserved.

In the above, I have assumed that your manifolds are Hausdorff. The same (counter)example works even if your manifolds are only locally Hausdorff – but a slightly more sophisticated argument is needed there.

$\endgroup$
1
$\begingroup$

In the meantime I have come up with a similar answer to Zhen Lin's. Consider $\mathbb{Q}$ with the discrete topology, then the colimit of the coproduct map $\mathbb{Q} \rightrightarrows \mathbb{R} \coprod \mathbb{R}$, where the two arrows are the two obvious inclusions, has the colimit $$ \begin{array}{rcl} \mathbb{R} \coprod \mathbb{R} & \to & \mathbb{R} \\ x & \mapsto & x, \end{array} $$ because any smooth function on $\mathbb{R}$ is determined by its values on $\mathbb{Q}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .