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Given $f$ and $g$ continuous maps from $X$ into $\mathbb{R}^{2}$, how to show that the straight line homotopy map $F(x,t)=(1-t)f(x)+tg(x)$ is continuous?

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2 Answers 2

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If $M:\mathbb R^3\to \mathbb R^2$ is the map given by $(x,y,t)\mapsto (tx,ty)$ and $A:\mathbb {R^2\times R^2}\to \mathbb R^2$ is the map given by $((x_1,y_1),(x_2,y_2))\mapsto (x_1+x_2,y_1+y_2)$ then $M$ and $A$ are continuous maps. Also the maps $I:\mathbb {R\to R}$ given by $t\mapsto t$ and $S:\mathbb{\to R}$ given by $t\mapsto (1-t)$ are continuous.

Then you can see that your map $F$ is given by $$(x,t)\mapsto((x,t),(x,t))\mapsto(f(x),(1-t),g(x),t)\mapsto((1-t)f(x),tg(x))\mapsto(1-t)f(x)+tg(x)$$

Where the second map is $f\times S\times g\times I$, the thirsd map is $M\times M$ and the fourth map is $A$. I hope you are convinced that the first map is continuous.

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If both $f$ and $g$ are continuous then $F$ is a composition of continuous functions which is also continuous.

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    $\begingroup$ What is the composition? $\endgroup$
    – TuoTuo
    Jan 22, 2016 at 1:49
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    $\begingroup$ @TuoTuo Well, he had just a little lapsus, it is the sum* :) $\endgroup$
    – user258738
    Jan 17, 2019 at 6:04

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