Let $M$ be a manifold, and $U$ be the forgetful functor from the category of Lie groups to the category of manifolds. My question is whether there is a universal arrow $(G, i)$ from $M$ to $U$? More precisely, whether there is a Lie group $G$ and a smooth map $i: M \rightarrow U(G)$ such that, for all smooth map $\phi: M \rightarrow U(Q)$, there exists a unique Lie group homomorphism $f: G \rightarrow Q$ with $U(f)\circ i = \phi$?


should we impose any conditions on $\phi$ to make it true?


An intuition to this question may make it interesting is that when we define a smooth structure on a manifold, whether we have identified a "relations" or "connections" between elements of this manifold simultaneously, such that "relations" take the form of Lie group. When we have a smooth map between manifolds, it transforms or collapses "relations" of one to another, and it collapses one Lie group to another.



Existence of universal arrow from manifolds to forgetful functor of Lie groups
Such arrow exists if and only if $Dim(M)=0$. In this case $G$ is the free group $F(M)$, considered as a $0$-dimensional manifold.


If $Dim(M)>0$, you can cook up a smooth map $M\rightarrow {\mathbb R}^n$, such that $\{ \alpha e\_k | \alpha \in (-1,1)\}$ is in the image for each standard basis element $e\_k$. This means that the corresponding Lie group homomorphism $G\rightarrow {\mathbb R}^n$ must be surjective. Thus, $Dim(G)\geq n$ for each $n$. So $G$ is not a Lie group.


Besides the smooth sets mentioned by Pavlov, you can use the category of Tychonoff topological spaces. There again the usual free group will give you the universal arrow. It is explained, for instance, in Abstract Harmonic Analysis by Hewitt and Ross.


