Contractible space as Codomain implies any two maps Homotopic

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Recall that a space Y is contractible if the identity map \text{id}_Y is homotopic to a constant map. Let Y be contractible space and let X be any space. Then, for any maps f,g: X\to Y, f\simeq g.

Proof: Let Y be a contractible space and let X be any space. \text{id}_Y\simeq c, where c is a constant map. There exists a map F: Y\times [0,1]\to Y such that F(y,0)=\text{id}_Y(y)=y, for y\in Y. F(y,1)=c(y)=b for some point b\in Y.

Let f,g: X\to Y be any two maps. Consider G:X\times [0,1]\to Y where G(x,t)=\begin{cases}    F(f(x),2t),&\ \ \ \text{for}\ 0\leq t\leq 1/2\\    F(g(x),-2t+2),&\ \ \ \text{for}\ \frac 12<t\leq 1    \end{cases}

When t=\frac 12, F(f(x),1)=b, F(g(x),1)=b. Therefore G is cts.

G(x,0)=F(f(x),0)=f(x),

G(x,1)=F(g(x),0)=g(x).

Therefore f\simeq g.

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Author: mathtuition88

Math and Education Blog

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