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.


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.

  • $\begingroup$ i have solved it but unable to show that the image is path connected $\endgroup$
    – george
    Apr 26 '15 at 0:27
  • $\begingroup$ @george The above argument shows the image is path connected by the definition of path connected with $f\circ\gamma$ the desired path. $\endgroup$
    – Eoin
    Apr 26 '15 at 0:44
  • 1
    $\begingroup$ yeah i realised thank you for your help $\endgroup$
    – george
    Apr 26 '15 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 $y_1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.