# The image of a path-connected set under a continuous map is path-connected

Show that if $X$ is path-connected and $f:X\to Y$ is a continuous map, then the image $f(X)$ is path-connected.

In order to show this is path connected I know the definition is :

Definition: A topological space $X$ is path connected if $\forall x,y \in X \, \, \exists$ continuous function $\gamma: [0,1] \rightarrow X$ such that $\gamma(0)=x$ and $\gamma(1) = y$. i.e. any two points can be connected with a continuous path.

## 2 Answers

Suppose $x,y\in f(X)\subset Y$ so that we can say $x=f(a)$ and $y=f(b)$. Then there is a path $\gamma:[0,1]\rightarrow X$ such that $\gamma(0)=a$ and $\gamma(1)=b$. The composition, $f\circ \gamma$ is then a continuous path on $f(X)$ with the desired property.

• i have solved it but unable to show that the image is path connected Apr 26, 2015 at 0:27
• @george The above argument shows the image is path connected by the definition of path connected with $f\circ\gamma$ the desired path.
– Eoin
Apr 26, 2015 at 0:44
• yeah i realised thank you for your help Apr 26, 2015 at 0:49

Let $y_0 = f(x_0), y_1 = f(x_1)$ two elements of $f(X)$.

Consider a path joining $x_0$ to $x_1$: $$\gamma \in C([0, 1], X);\\ \gamma(i) = x_i\ \ \ \ (i\in \{0, 1\})$$

Then $f \circ \gamma$ is a path joining $y_0$ and $y_1$.