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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ yeah i realised thank you for your help $\endgroup$ – george Apr 26 '15 at 0:49
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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 $y_1$.

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