Does a continuous mapping $f : E \subset \Bbb R^{n} \to \Bbb R^{m} $ preserve connectedness? If $E $ is a connected subset of $\Bbb R^{n} $, does a continuous mapping $f(E)$ into $\Bbb R^{m}$ preserve connectedness?
I think yes by definition of continuity since otherwise there would be some point s.t. $f(E)$ would be a disconnected set, but this point would be a discontinuity
 A: Hint: Startby assuming $f: M \to N$ is continuous and surjetive and $M$ is connected. Suppose $N = A \cup B$, with $A,B$ disjoint open subsets of $N$. Now since $M$ is connected, what can you say about $f^{-1}(A)$ or $f^{-1}(B)$ where $M = f^{-1} (A) \cup f^{-1} (B)$? 
Conclude that, for any $E \subseteq M$ connected, $f: E \to f(E)$, takes $E$ onto a connected subset of $N$.  
A: The answer is yes and holds for the more general case that $E$ is any connected space. We prove this by proving the equivalent (contrapositive):

Let $f: X \rightarrow Y$ be continuous. If $f(X)$ is disconnected, then $X$ is disconnected.

Suppose that $f(X) = A \cup B$ is a separation, i.e. $A, B \subset f(X)$ is open, $A, B \neq \varnothing, A \cap B = \varnothing, A \cup B = f(X)$.
Let $C = f^{-1}(A), D = f^{-1}(B)$. Then $C, D \neq \varnothing, C \cap D = \varnothing$.
Write $A = f(X) \cap U$, where $U \subset Y$ is open.
Then $C = f^{-1}(A) = f^{-1}(U)$ is open, since $f$ is continuous.
Similarly, $D$ is open, and hence $X$ is disconnected.
A: Suppose $\;f(E)\;$ isn't connected $\;\iff\;$ there exists a continuous onto map $\;h: f(E)\to\{0,1\}\;$ , the last two-elements set with the inherited euclidean topology from $\;\Bbb R\;$ (and thus discrete), but then:
$$E=(h\circ f)^{-1}(\{0\})\cup(h\circ f)^{-1}(\{1\})$$
and since both $\;(h\circ f)^{-1}(\{0\})\;,\;\;(h\circ f)^{-1}(\{1\})\;$ are open and non empty (why and why?), $\;E\;$ isn't connected.
