Smooth circle action, $\chi(M^{S^1}) = \chi(M)$

I need a reference for the following result (which I can prove myself, but my proof is rather ugly and I would prefer to just cite the statement instead or re-proving it):

Let $M$ be a closed smooth manifold and let $S^1$ act smoothly on $M$. Then the set of fixed points of the action, $M^{S^1}$, is again a (not necessarily connected) smooth manifold and $\chi(M^{S^1}) = \chi(M)$.

• A G-manifold has a G-CW complex structure, and for G-cells other than the discs with trivial $S^1$ action the Euler characteristic is zero. – Mike Miller Dec 16 '16 at 1:22

1. Take a generator $g$ of the group $S^1$. Then the rigidity of Euler characteristic number tells us $$\chi(M)=\sum_i(-1)^i\mathrm{Tr}[g|_{H^i(M)}]$$
2. Applying the Atiyah-Bott fixed point formula, we get $$\sum_i(-1)^i\mathrm{Tr}[g|_{H^i}(M)]=\int_{M^{S^1}}e(M^{S^1}),$$ where $e(M^{S^1})$ is the Euler characteristic class.
Consider a tubular neighborhood $T$ of the fixed set (submanifold) $M^{{\mathbb S}^1}$. By Mayer-Vietoris sequence, relatively to the open covering $\{U = M-{\mathbb S}^1, T\}$ of $M$, we get $$\chi (M) = \chi (U) + \chi(T) - \chi(U\cap T).$$ But, $\chi (U) =\chi(U\cap T)=0$ since the circle action is free on $U$. Finally, since $T$ equivariantly retracts on $M^{{\mathbb S}^1}$ then $$\chi (M) = \chi (T) = \chi(M^{{\mathbb S}^1}).$$