The following proof of
An arbitrary product $X=\prod_{\alpha \in J} X_\alpha$ is connected iff $X_{\alpha }$ is connected for each $\alpha\in J$ is provided by
my professor.I've some queries in this proof.
Let X is connected then we shall show that each $X_{\alpha}$ is connected,where $\alpha\in J,$some index set.
Consider $\pi_{\alpha}:X\rightarrow X_{\alpha},{\alpha}\in J$.
Since $\pi_{\alpha}$ is a projection map so it is continuous & $X$ is connected by assumption.
Using the result:"Continuous image of a connected set is connected."
we have $X_{\alpha}$ is connected.
Conversely,
Result Used:
X is connected iff every continuous function $f:X\rightarrow ${$0,1$} is constant.
Fix $f:X\rightarrow ${$0,1$} a continuous function.
Fix $(a_{\alpha})\in X$ and for each $\alpha\in J,$define $\varphi_{\alpha}:X_{\alpha}\rightarrow X$ by $\varphi_{\alpha}o\pi_{\beta}(x) = \begin{cases} a_{\beta}, & \text{if $\beta\neq \alpha$} \\[2ex] x, & \text{if $\beta=\alpha$ } \end{cases}$.
Let $g_{\alpha}=fo\varphi_{\alpha}:X_{\alpha}\rightarrow${$0,1$} for each $\alpha\in J.$Then each $g_{\alpha}$ is constant since $X$ is connected.
Hence,for each $\alpha\in J, f$ is constant on $\varphi [X_{\alpha}].$
Say,Without loss of generality,for some $\alpha_0\in J,f[\varphi_{\alpha_0[X_{\alpha}]}]=${$0$}.Since,$(a_{\alpha})\in \varphi_{\beta}[X_{\beta}]$ for each $\beta \in J,$ it must be that $\forall \alpha\in J,f[\varphi_{\alpha_0[X_{\alpha}]}]=${$0$}.
Then $f^{-1}$[{$0$}] is an open subset of $X$ by continuity which contains all of these elements.$\tag{*}$
1.(Please clarify how $f^{-1}$[{$0$}] is open subset of $X$ :Is it because {0} is open in {0,1},$f$ is continuous,so inverse image of open set is open? )
Let $U\subset f^{-1}$[{$0$}] be a basis element of $X$ containing $(a_{\alpha})$ so that $\pi_{\alpha}[U]=X_{\alpha}$ for all but finitely many $\alpha \in J$.Let K be the finite subset of $J$ comprised of these $\alpha $ and let $(x_{\alpha})\in U$ be that element such that $x_{\alpha}=a_{\alpha} $ if $\alpha \in K.$
Changing the coordinates of $(x_{\alpha})$ indexed by $K$ one at a time,it follows from ($*$) that each time we change it,it remains in $f^{-1}$[{$0$}].Since $\pi_{\alpha}[U]=X_{\alpha}$ for all but finitely many $\alpha \in J$.
We have shown that $\prod_{\alpha \in J} X_\alpha=f^{-1}$[{$0$}].(##2.I'm not getting this sudden transition.Please explain!!)
If there is some scope of refinement in the above proof,feel free to express your views.Also check this proof critically...