I have to solve the following exercise:
Let $S^m(r)=\{x\in \mathbb{R}^{m+1}|\sum_i (x^i)^2=r^2, r>0\}$ and let $\Phi$ be the Stereographic projection $$\Phi: S^m(r)\setminus\{(0,...,0,r)\}\rightarrow \mathbb{R}^m$$ Show, $\Phi$ is a conformal mapping, i.e. for $(S^m,\tilde{g})$ and $(\mathbb{R}^m,g)$ there exists $f\in C^2(S^m(r))$ with $f>0 $ such that $$f^2\tilde{g}=\Phi^*g$$ where $\tilde{g}$ is the induced metric $S^m\subset \mathbb{R}^{m+1}$ and $g$ the Euclidean metric.
Now, I already have made the connection between the metrics in this way: \begin{align*} \tilde{g}=g^{sphere}_{\alpha\beta}=&\sum_{i,j}\frac{\partial}{\partial x^\alpha}X^i\frac{\partial}{\partial x^{\beta}}X^jg_{ij}\\ =& \frac{1}{\partial x^\alpha \partial x^{\beta}}(\sum_i \partial^2 X^i)^2 \end{align*}
If I'm comparing now $\tilde{g}$ and $g$ there is of course the difference of $\sum_{i,j}\frac{\partial}{\partial x^\alpha}X^i\frac{\partial}{\partial x^{\beta}}X^j$ between them, for which I can reason, the properties mentioned for $f$, but is that enough?
Don't I need to look at some explicit expression of $\Phi$ to compare them? Otherwise any map between a sphere with induced metric and Euclidean space would be conformal, which is not the case, right?