Proposition 3.2.2.label Let $X,Y$ be topological such that $X$ is connected then for any $f\in C(X;Y)$, $f(X)$ is connected w.r.t. subspace topology.
Proof. Let $\curl{U,V}\suf \cali{P}(Y)$ be an open cover of $f(X)$ in the sense that $f(X)\suf U\cap V$. By continuity of $f$ and connectedness of $X$ we have $f\inv(U)\cap f\inv(V)=f\inv(U\cap V)\neq\emp\implies U\cap V\neq\emp$.$\square$
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