A problem about connected components‘ image under onto and continous map Here is a problem on Topoloy written by Marco Manetti, page89, question 5.4.

Let $f:X \rightarrow Y$ be an identification. Show that if the connected components of $X$ are open, so are the connected components of $Y$.

I think that if we can prove that $A$ is a connected components of $Y$ implies $f^{-1}(A)$ can written as several connected components of $X$, then this problem can be solved. When I tried to prove it, I found that even if $f$ is just a onto and continous, this conclusion still holds. Here is my proof. 

If $B\subset X$ is a connected component, since $f$ is continous, then $f(B)$ is connected, hence contained in some connected component of $Y$. Because $f$ is onto, each pre-image of connected components of $Y$ can be written as several connected components of $X$.

My question is, can we replace the identification by onto and continous in this problem?

The answer is no, the claim is not true if you replace $f$ by continuous and onto. 
The problem is the fact that you still need to show that the image of open is open, continuity does not satisfies.
Consider the following example: pick $Y$ to be any topological space whose components are NOT open. Pick $X$ to be the same set, equipped with the discrete topology. 
Define $f :X \to Y$ to be $f(x)=x$. Then $f$ is continuous, onto, the components of $X$ are open, but the components of $Y$ are not.
P.S. An alternate approach for the problem is the following: Let $\{ C_i \}_{i}$ be the components of $X$ and $\{ D_j \}_j$ the components of $Y$. 
You know that for each $i$ there exists some $j$ such that $f(C_i) \subset D_j$. Use this together with ontoness to show that for each $j$ you have 
$$f^{-1}(D_j) = \bigcup_{f(C_i) \subseteq D_j} C_i $$
