The number of connected components of X is greater than number of connected component of Y If X and Y are topological spaces such that there exist a continuous surjection from X to Y. Then the number of connected components of X is greater than equal to that of Y. I can see this logically. But how should I begin to prove it. I mean what am I to show?. If anyone can suggest to me what should I try and show then it will be of great help.
 A: Let $f : X \to Y$ be continuous and let $A$ be a connected subset of $X$. Then the image $f[A]$ of $A$ under $f$ is also connected. (If you haven't already seen a proof of this, note that if $f[A]$ is disconnected by open subsets $B$ and $C$ of $Y$, i.e., if $f[A] = (f[A] \cap B) \cup (f[A] \cap C)$, where $B$ and $C$ are both open in $Y$ and $(f[A] \cap B) \cap (f[A] \cap C) = \emptyset$, then $A = (A \cap f^{-1}[B]) \cup (A \cap f^{-1}[C])$ gives a disconnection of $A$ by the open subsets $f^{-1}[B]$ and $f^{-1}[C]$ of $X$.)
Hence a continuous $f$ maps each connected component of $X$ into a single connected component of $Y$ (since the connected components are the maximal connected sets and $f[A]$ is connected for each connected component $A$ of $X$ by the above remark). So $f$ induces a map, $f^*$ say, from the set, $X^*$ say, of connected components of $X$ to the set, $Y^*$ say, of connected components of $Y$. If $f$ is surjective then so is $f^*$ (do you see why?). If there is a surjection from $X^*$ onto $Y^*$, then $|X^*| \ge |Y^*|$
