If $f\colon X \to Y$ is continuous, and the $U$ is the subset connected of $X$, then $f(U)$ is the subset connected of $Y$.

Asked Nov 15 '22 at 05:14

Active Nov 16 '22 at 19:09

Viewed 50 times

I want to prove this part.

Let $f:X \rightarrow X_2$ be a continuous function from $(X_1, \tau_1)$ onto $(X_2, \tau_2)$. If $U$ is connected subset of $X_1$ then $f(U)$ is connected subset of $X_2$.

By contradiction. Suppose that $f(U)$ is disconnected subset of $X_2$. Then there exists two nonempty open sets $A$ and $B$ such that $A\cap B=\emptyset$, $A\cup B= f(U)$. Let $C=f^{-1}(A), D=f^{-1}(B)$. And to prove that C and D are two nonempty disjoint sets, which is contradiction.

Is this one correct?

edited Nov 16 '22 at 19:09

Arturo Magidin

398,050

asked Nov 15 '22 at 05:14

lulu

2

Does this answer your question? If $f: X \rightarrow Y$ is continuous, and $X$ is connected, then $f(X)$ is connected – azif00 Nov 15 '22 at 05:18
@azif00 Thanks for the line. Since I am not to much sure is that the same problem. Since U is connected subset, not to say $x_1$ is connected. – lulu Nov 15 '22 at 05:46
1

Your welcome :) And it is the same thing: the restriction $f|_U \colon U \to X_2$ of $f \colon X_1 \to X_2$ to $U$ is continuous and its image is $f(U)$, so, by the link, $f(U)$ is connected since $U$ is. – azif00 Nov 15 '22 at 05:51

If $f\colon X \to Y$ is continuous, and the $U$ is the subset connected of $X$, then $f(U)$ is the subset connected of $Y$.

0 Answers0