# Image of continous function for which $f(c,0) = (0,0)$ is connected

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?

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.
$$f(X)$$ 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.
• Not sure I understood. If for example $A = [2,3]$, then $X = [2,3] \times [0,1]$ forms a squared region on the plane?
• @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? May 6 at 23:19