My problem is in analytic manifolds.According to Cohn's book a function $f$ in a manifold $M$ is analytic at $p \in M$ if it can be expressed as a power series of $\sigma(p)=(x_{0})$. That means $f=f(x)= c_{0}+c_{1}(x-x_{0})+c_{2}(x-x_{0})^{2}+\cdots $.Here $x$ is the image of points in neighborhood of $p$ under the chart $\sigma$
$g$ be an analytic function at $f(p)$. Hence $g = d_{0}+d_{1}(y-y_{0})+d_{2}(y-y_{0})^{2}+ \cdots$ where $\tau(f(p))=(y_{0})$ and $y$ is the image of points in neighborhood of $f(p)$ under the chart $\tau$ .So $y= f(x)$
Here $\sigma$ be a chart at $p$ and $\tau$ be a chart at $f(p)$. Composition of charts is again a chart. So I want to show that $g \circ f$ can be expressed as a power series at $\sigma(p)=(x_{0})$.
$g(f(p))= d_{0}+d_{1}(y-y_{0})+d_{2}(y-y_{0})^{2}+ \cdots $.(x)$
Since I am working in analytic manifolds any two charts in $M$ are analytically related.That is same as saying $\phi = \tau \circ \sigma^{-1}$ can be expressed as a power series at $\sigma(p)$.
If I can claim that $g \circ f$ can be expressed as a power series in $p$ I can complete the proof.
Now what I need is I want to replace $y$ with $x$.I think I can do it since $f(x)=y$ and then I want to replace $y_{0}=\tau(f(p))$ by $\sigma(p)=x_{0}$. Then proof will be complete. Can any one help in this step ? I don't want to use any conditions for analyticity.