It is a consequence of Malgrange's preparation theorem for differentiable functions that $C^{\infty}(M)$ is a faithfully flat $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of analytic functions on $M$). See Corollary 1.12, Chapter VI of his book "Ideals of differentiable functions".

On the other hand $C^{\omega}(M)$ is a flat $C^{\omega}(N)$-module as the argument pointed out by Greg Stevenson shows.

I believe that these two facts can be put together to give a positive answer to the question.

It is a consequence of Malgrange's preparation theorem for differentiable functions that $C^{\infty}(M)$ is a faithfully flat $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of analytic functions on $M$). See Corollary 1.12, Chapter VI of his book Ideals of differentiable functions.

On the other hand $C^{\omega}(M)$ is a flat $C^{\omega}(N)$-module as the argument pointed out by Greg Stevenson shows.

I believe, but don't know how, these two facts can be put together to give a positive answer to the question.

It is a consequence of Malgrange's preparation theorem for differentiable functions that $C^{\infty}(M)$ is a faithfully flat $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of analytic functions on $M$). See Corollary 1.12, Chapter VI of his book "Ideals of differentiable functions".

On the other hand $C^{\omega}(M)$ is a flat $C^{\omega}(N)$-module as the argument pointed out by Greg Stevenson shows.

I believe that these two facts can be putted together to give a positive answer to the question.

It is a consequence of Malgrange's preparation theorem for differentiable functions that $C^{\infty}(M)$ is a faithfully flat $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of analytic functions on $M$). See Corollary 1.12, Chapter VI of his book "Ideals of differentiable functions".

On the other hand $C^{\omega}(M)$ is a flat $C^{\omega}(N)$-module as the argument pointed out by Greg Stevenson shows.

I believe that these two facts can be put together to give a positive answer to the question.