Open connected components

Question

I don't know how solve this exercise: "If $f: X \rightarrow Y $ is an identification and the connected components of X are open show that the connected components of Y are open" My attempt: Consider $C$ connected components and consider $f^{-1}(C)$ how can I show that $f^{-1}(C)$ is union of connected components? From here $C$ is open because $f$ is an identification.

Answer 1

Let $C$ be a connected component of $Y$. Consider $x\in f^{-1}(C)$ and let $B_x$ be a connected component of $x$ in $X$. Since continous functions map connected sets to connected sets, then $f(B_x)\subseteq C$. Thus $B_x\subseteq f^{-1}(C)$. Since $x$ was picked arbitrarly then

$$f^{-1}(C)=\bigcup_{x\in f^{-1}(C)} B_x$$

and since each $B_x$ is open by assumpion, then $f^{-1}(C)$ is open and so is $C$.