I am reading Orbifolds as Groupoids: an Introduction https://arxiv.org/abs/math/0203100
Let $\mathcal{G}$ be a Lie groupoid. A right $\mathcal{G}$ space is a smoooth manifold $E$ equipped with an action by $\mathcal{G}$. Such an action is given by two maps $\pi:E\rightarrow \mathcal{G}_0$ and $\mu: E\times_{\mathcal{G}_0}\mathcal{G}_1\rightarrow E$. The latter map is defined on pairs $(e,g)$ with $\pi(e)=t(g)$, and written $\mu(e,g)=e.g$. It satisfies the usual identities for an action, viz. $$\pi(e.g)=s(g), e.1_x=e, (e.g).h=e.gh$$for $z\xrightarrow{h}y\xrightarrow{g}x$ in $\mathcal{G}$ and $e\in E$ with $\pi(e)=x$.
This being said, he defines what is called pull back of $\mathcal{G}$ space.
If $\phi:\mathcal{H}\rightarrow \mathcal{G}$ be a morphism of groupoids, there is an obvious functor $$\phi^*:(\mathcal{G}-\text{spaces})\rightarrow (\mathcal{H}-\text{spaces})$$ mapping $E$ into the pullback $E\times_{\mathcal{G}_0}\mathcal{H}_0$ with the induced action.
I understand how they have defined the space but could not see how this would be a $\mathcal{H}$ space.
Some one in their notes said,
There is an obvious action of $\mathcal{H}_1$ on $E\times_{\mathcal{G}_0}\mathcal{H}_0$ and we write $\phi^*E$ for the resulting space.
I am not able to guess what is that obvious action is.
We do have a map $\pi^*:E\times_{\mathcal{G}_0}\mathcal{H}_0\rightarrow \mathcal{H}_0$.
To see that $E\times_{\mathcal{G}_0}\mathcal{H}_0$ is a $\mathcal{H}$ space, we need to give a map $(E\times_{\mathcal{G}_0}\mathcal{H}_0)\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow E\times_{\mathcal{G}_0}\mathcal{H}_0$. If we have $$\mu^*((e,a),h)=(e',b)$$ then, for this to give an action map, we need to have (from definition)
$\pi^*(\mu^*((e,a),h))=s(h)$
As we have assumed $\mu^*((e,a),h)=(e,b)$ we need to have $\pi^*(e,b)=s(h)$ i.e., $b=s(h)$. So, $$\mu^*((e,a),h)=(e',s(h))$$ We need to fix some thing for $e'$ as well. For $(e',s(h))\in E\times_{\mathcal{G}_0}\times \mathcal{H}_0$, we need to have $\pi(e')=\phi(s(h))$. Thus, if we have to define a map $\mu^*:(E\times_{\mathcal{G}_0}\mathcal{H}_0)\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow E\times_{\mathcal{G}_0}\mathcal{H}_0$ that makes $E\times_{\mathcal{G}_0}\mathcal{H}_0$ to be a $\mathcal{H}$ space, it has to be of the form $$\mu^*((e,a),h)=(e',s(h))$$ where $e'\in E$ is such that $\pi(e')=\phi(s(h))$. Now the problem is, how would I know such $e'$ exists?
$e$ is not an obvious choice for $e'$ as we only have $\pi(e)=\phi(t(h))$ and can not say anything if it is same as $\phi(s(h))$.
Any comments are welcome.
EDIT : Can some one tell me if you think $\pi:E\rightarrow \mathcal{G}_0$ should be assumed to be a submersion? I am not sure otherwise how would one get a smooth manifold structure on $E\times_{\mathcal{G}_0}\mathcal{G}_1$. If $\pi:E\rightarrow \mathcal{G}_0$ is a submersion, then I say that these two maps $\pi,\phi$ intersect transversally, so by transversality theorem, there would be a smooth manifold structure on the pull back $E\times_{\mathcal{G}_0}\mathcal{G}_1$... I do not know how to see if this is not the case.
The idea is that $\mathcal{H}$ acts through $\mathcal{G}$, so the obvious candidate is $e' = \mu(e, \phi(h))$. You can easily check the required properties.