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Nov 7, 2011 at 14:41 history edited Jeff Strom CC BY-SA 3.0
changed $g(y)$ to $g(x)$
Nov 7, 2011 at 13:19 vote accept Chulumba
Nov 7, 2011 at 11:45 comment added Buschi Sergio In your example you make merely a funtion $F$ such that for any $t$ the function $F_t: x \mapsto F(x, t)$ is contnuous, of course this is too weak for have the (global) continuity of $F$. Anyway if you parametrize the paths by the homotopy contraction maps you get a (continuous) homotopy $F$ from $f$ to $g$. Let $G: X \times I\to Y: 1_Y\ \tilde\ \ C(y_0)$ (contraction homotopy from identity map to costant map on the point $y_0$). then join the maps $F(x, t):= G(f(x), 2t)\ t\in[0,1/2]$ and $F(x, t):= G(g(x), 2-2t)\ t\in[1/2,1]$ .
Nov 7, 2011 at 10:47 answer added Martin Brandenburg timeline score: 3
Nov 7, 2011 at 10:38 history asked Chulumba CC BY-SA 3.0