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Suppose $\alpha, \beta : I \to X$ are paths and suppose $\alpha $ is homotopic to $\beta$, $\alpha \cong \beta$. So, can find a continuous function $F(s,t) = f_t(s)$ such that

$$ f_t(0) = \alpha(0) = \beta(0) = x_0 , \; \; \; f_t(1) = \alpha(1) = \beta(1) = x_1$$

Also, we know $f_0(s) = \alpha(s) $ and $f_1 = \beta$. We want to show $\beta \cong \alpha$. If we put $F(s, 1 - t) = f_{1-t}(s)$, then

$$ f_0(s) = \beta(s), \; \; \; \; f_1 = \alpha$$

so, $\beta \cong \alpha$ IS this correct? Any feedback?

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    $\begingroup$ @StefanH Although the questions are similar, this is asking for feedback and so cannot be a duplicate. $\endgroup$
    – user1729
    Commented Nov 10, 2013 at 15:20

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The answers at Symmetry of "is homotopic to" detail in the proof give detailed proofs on why $H(x,t)=F(x,1-t)$ is continuous. The basic idea is that $H$ is the composition $$I×I⟶I×I\longrightarrow X$$ where the first arrow is the map $(x,t)↦(x,1-t)$ and the second map is $F$. The first map again is continuous because its coordinates are.

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