# Lie groupoids being homotopy equivalent

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?

• Thank you @YCor – Praphulla Koushik Feb 22 at 13:22
• There might be more than one: do you want to consider that equivalences of groupoids are homotopy equivalences, or you want to keep the information of the space of objects ? – Simon Henry Feb 22 at 17:18
• @SimonHenry I want to keep information on the space of objects.. I do not know if it is reasonable to call equivalence of Lie groupoids as Homotopy equivalence – Praphulla Koushik Feb 23 at 1:39
• Another source to learn about when two Lie groupoids are homotopy equivalent is the homotopy theory created by Joost Nuiten: link.springer.com/article/10.1007/s10485-019-09563-z. If I had more time, I'd try to dig into the paper to find the weak equivalences, but I'm swamped. So I just leave you the reference. – David White Mar 3 at 0:08

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.

• I think I am missing something here...You are only saying about Homotopy equivalence of topological spaces... There is a +1 to your answer, so, there must be something interesting in your answer, it is just that I am misunderstanding – Praphulla Koushik Feb 23 at 1:36
• it might be confusing because it's formally identical to "homotopy equivalence" of topological spaces. But in the above all the objects are topological groupoids and all maps between them are maps of groupoids. Note that a topological space gives rise to a pretty canonical topological groupoid by declaring that there are no non-identity morphisms, i.e. the space $X$ becomes the groupoid $X\rightrightarrows X$. – John Greenwood Feb 23 at 2:22
• Yes, that one I am aware of associating a topological groupoid for a topological space, Lie groupoid for a manifold.. Has this formal notion of Homotopy equivalence used any where else in literature, in similar way a notion of homotopy equivalence of topological spaces used? – Praphulla Koushik Feb 23 at 2:59
• Yeah, this is the standard notion of homotopy for groupoids. In fact this construction using the "cylinder object" $X\times[0,1]$ is the standard way to define "homotopy" in any model category see pg 233-234 of Quillen's "Rational Homotopy theory" – John Greenwood Feb 23 at 19:28
• I believe that is the standard way to talk about Homotopy equivalences. But, I was thinking if we should consider the category of Lie groupoids whose morphisms are morphisms of Lie groupoids or that of bibundles (I have the set up of differentiable stacks in mind).. So, in that case should we involve the notion of bibundle (generalized morphism) when defining the notion of Homotopy equivalence? – Praphulla Koushik Feb 24 at 6:44