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Let $A \subset \mathbb{R}$ and $X = A \times [0,1] \subset \mathbb{R}^2$. Let $f : X \to \mathbb{R}^2$ be continous function for which $f(c,0) = (0,0)$ for all $c \in A$. Show that the image $f(X)$ is connected.

I'm trying to show that this is path-connected, but bit stuck. I was instructed to pick $(a,b),(c,d) \in X$ and then construct a path to $(a,0)$ and $(c,0)$, but not sure what this means?

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Let $A_i$ be the connected components of A. Then, since $A=\bigcup_i A_i$, we have $$ X=\bigcup_i A_i \times [0,1] ; \quad f(X)=\bigcup_i f(A_i \times [0,1]) $$

Then, by continuity $f(A_i\times[0,1])$ is connected. Since they intersect at $(0,0)$ by a certain topological lemma (in Spain we call it the hanger's lemma: Proof here), $f(X)$ is connected.

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$f(X)$ is the union of the sets $\{f(S_c):c \in A\}$ wheer $S_c=\{c\}\times [0,1]$. $f(S_c)$ is connected for each $c$ and these set have $(0,0)$ in common. Hence the union is connected.

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  • $\begingroup$ Not sure I understood. If for example $A = [2,3]$, then $X = [2,3] \times [0,1]$ forms a squared region on the plane? $\endgroup$
    – Timo
    May 6 at 14:28
  • $\begingroup$ @Timo Any point in $f(X)$ is of the form $f(c,t)$ for some $c \in A, t \in [0,1]$. Observe that $f(c,t)$ belongs to $f(S_c)$ with my defintion of $S_c$. Do you know that union of connected sets having at least one point in common is always connected? $\endgroup$ May 6 at 23:19

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