Diffeomorphism group of product manifold

Question

For a given differentiable manifold $M$, the diffeomorphism group $\mathrm{Diff}\left( M \right)$ of $M$ is the group of all $C^\infty$ diffeomorphisms of $M$ to itself. Consider a product manifold of the form $M \times N$. My question is: is $\mathrm{Diff}\left( M \times N\right) \cong \mathrm{Diff}\left(M\right) \times \mathrm{Diff}\left( N\right)$?

My (physicist's) intuition is no, for consider $\mathbb{R}^2 \cong \mathbb{R} \times \mathbb{R}$. It seems like $\mathrm{Diff}\left(\mathbb{R}\right) \times \mathrm{Diff}\left( \mathbb{R}\right)$ on $\mathbb{R}^2$ can give smooth coordinate transformations along two 'axes', but it can't give 'twists' etc., as could $\mathrm{Diff}\left(\mathbb{R}^2\right)$.

Am I right? And, if so, is anyone able to help make my intuition more precise? Thanks for any help!

Answer 1

They are not isomorphic, Suppose that $M$ and $N$ are finite of cardinal respectively $m$ and $n$ $Diff(M)$ is the symmetric group $S_m$, $Diff(N)=S_n$ and $Diff(M\times N)=S_{mn}$.

Answer 2

As the other comment mentions, the answer is no. For example, even if you want to consider only those diffeomorphisms which preserve one of the projections $X \times Y \to X$ we have the space of maps $X \to \text{Diff}(Y)$. (i.e. fiber transformations) The homotopy groups of $\text{Diff}(X\times Y)$ may not even be the same as those of $\text{Diff}(X)\times \text{Diff}(Y)$. See, for example, the Gluck twist: https://www.jstor.org/stable/1993581?seq=1#page_scan_tab_contents