Let $M,N$be two smooth manifolds. Let $f,g:M\rightarrow N$ be two smooth maps. We have the notion of a homotopy (smooth homotopy) from the maps $f$ to the map $g$.


Is there a similar concept for morphisms of Lie groupoids?


Suppose $\mathcal{G}=(\mathcal{G}\_1\rightrightarrows \mathcal{G}\_0)$ and $\mathcal{H}=(\mathcal{H}\_1\rightrightarrows \mathcal{H}\_0)$ be Lie groupoids.


Let $(\phi\_1,\phi\_0),(\psi\_1,\psi\_0): \mathcal{G}\rightarrow \mathcal{H}$ be two morphisms of Lie groupoids. Is there any notion of a “homotopy” from $(\phi\_1,\phi\_0)$ to $(\psi\_1,\psi\_0)$?


Further, is there a notion of when two Lie groupoids are homotopy equivalent?


Is there a notion for topological groupoids?



Lie groupoids being homotopy equivalent
Yes there is! Here is one way to go.


If $X=(X\_{1}\rightrightarrows X\_{0})$ is a topological groupoid, then $X\times [0,1]=(X\_{1}\times[0,1]\rightrightarrows X\_{0}\times[0,1])$ is also a topological groupoid.


So the notion of homotopy is: if $f,f':X\rightarrow Y$ are two maps, then a homotopy between them is a map $F:X\times[0,1]\rightarrow Y$ that restricts to $f$ and $f'$ at $X\times\{0\}$ and $X\times\{1\}$. 


And as soon as you have the notion of a homotopy of maps, you have the notion of a homotopy equivalence, defined as usual to be a pair of maps $f:X\rightarrow Y$ and $g:Y\rightarrow X$ and a pair of homotopies between the composites $fg$ and $gf$ and the respective identity maps.


