Show that the projection $M_1\times \cdots \times M_n\to M_i$ given by $p_i((x_1,\cdots,x_i,\cdots,x_n)) = x_i$ is a continuous and open map. I have the following question:
Let $M_i$ be metric spaces. Show that the projection $M_1\times \cdots \times M_n\to M_i$ given by $p_i((x_1,\cdots,x_i,\cdots,x_n)) = x_i$ is a continuous and open map. Is $p_i$ an homeomorphism?
Well, I have the definition that a function $f$ is continuous iff  the inverse image of every open set is open. But... How?
 A: 1) A counterexample for the statement that $p_{i}$ is homeomorphism is as following:
Consider $p_{1}:\mathbb{R}^{2}\rightarrow\mathbb{R},(x,y)\mapsto x$
This map is not injective since $p_{1}(0,1)=0=p_{1}(0,0)$ hence cannot be a homeomorphism.
2) To show $p_{i}$ is continuous: Let $U_{i}$ be open in $M_{i}$, then $p_{i}^{-1}(U_{i})=M_{1}\times\cdots M_{i-1}\times U_{i}\times M_{i+1}\times\cdots M_{n}$. This will be open in the product topology of $M_{1}\times\cdots\times M_{n}$ since each components of the product $M_{1}\times\cdots M_{i-1}\times U_{i}\times M_{i+1}\times\cdots M_{n}$ is open in $M_{i}$.
This holds even for $M_{i}$ are not metric spaces, as long as $M_{1}\times\cdots\times M_{n}$ is given the product topology.
3) To show $p_{i}$ is open mapping, we use the metric space condition here:
Denote the metric on $M_{i}$ to be $d_{i}$.
Given $U$ open in the $M_{1}\times\cdots\times M_{n}$. Consider some point $x_{i}\in p_{i}(U)\subset M_{i}.$ We know there exists some point $y=(x_{1},\cdots,x_{i},\cdots,x_{n})\in U$. Since $U$ is open, we know there exist some $\delta>0$ such that if $z=(z_{1},\cdots,z_{i},\cdots,z_{n})$ satisfies $$(d_{1}(z_{1},x_{1})^{2}+\cdots+d_{i}(z_{i},x_{i})^{2}+\cdots+d_{n}(z_{n},x_{n})^{2})^{\frac{1}{2}}<\delta,$$
then $z\in U$.
Now can you show that the $\delta$-neighborhood of $x_{i}$ is contained in $p_{i}(U)$?
