Path Connectivity of the Range of Path Connected Sets: Cts. Functions

# Path Connectivity of the Range of a Path Connected Set under Continuous Functions

Suppose that we have two topological spaces $X$ and $Y$, and that $f : X \to Y$ is continuous. If $X$ is a path connected space, then we should expect that the range, $f(X)$, is also path connected. If we take any two points $x$ and $y$ in the range, then there exists two points in the domain, $u$ and $v$ that are mapped to $x$ and $y$ respectively.

Now consider a path from $u$ to $v$ in $X$. Then if we take the image of all the points in this path, then since $f$ is continuous, the resulting image of points will be continuous and will be a path from $x$ to $y$.

We prove this result in the following theorem.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be continuous. If $X$ is path connected in then $f(X)$ is path connected in $Y$. |

**Proof:**Let $x, y \in f(X)$. Then there exists $u, v \in X$ such that $f(u) = x$ and $f(v) = y$. Since $X$ is path connected, there exists a path $\alpha : [0, 1] \to X$ such that $\alpha(0) = u$ and $\alpha(1) = v$.

- We define a path $\beta : [0, 1] \to f(X)$ from $x$ to $y$ as:

\begin{align} \quad \beta = f \circ \alpha \end{align}

- Then $\beta$ is continuous since $f$ and $\alpha$ are continuous. Furthermore, we see that:

\begin{align} \quad \beta(0) = f(\alpha(0)) = f(u) = x \end{align}

(3)
\begin{align} \quad \beta(1) = f(\alpha(1)) = f(v) = y \end{align}

- Thus $f(X)$ is path connected.