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Let $A = (C([0,1], \mathbb{R}), d_{\sup})$ be the space of continuous functions from $[0,1]$ to euclidean space $(\mathbb{R}, d_e)$. $d_{\sup}$ is supremum metric $$d_{\sup}(f,g) = \sup_{t \in [0,1]}\{|f(t)-g(t)|\}$$ ($|\dot{}|$ means absolute value)

Let $T, S$ be subspaces of $A$ $$T = \{f \in A: f(0) \neq f(1)\}$$ $$S = \{f \in A: f(0) ≤ f(1)\}$$ Check if $T, S$ are connected spaces and if are contractible spaces.

By definition, if $\operatorname{id}_T$ is homothopic with $\psi(f) = g$ where $g \in T$ ($g$ is constant) then $T$ is contractible (the same for $S$).

So lets choose $g = g(x):[0,1] \rightarrow \mathbb{R};\ g(x) = x$.

Define $H: (T \times [0,1]) \rightarrow T$ $$H(f, t) = (1 − t)f + tg$$ (multiplication function by the number defined as a new function such that $a\dot{}f = g \Longleftrightarrow \forall_{x\in\text{domain}} g(x)=af(x)$)

$H(f, 0) = (1-0)f+0g=f$ so $H(f,0)$ is $\operatorname{id}_T$

$H(f, 1) = (1-1)f+1g=g$ so $H(f,1)$ is $g(x)=x$

And now I should show that $H$ is continuous, however I have no idea how to show that (even I do not know if $H$ is continuous on $T$ or $S$). Or maybe there is better way to do this task?

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    $\begingroup$ $H$ is supposed to have values in $T$, which isn't the case in your definition. $\endgroup$
    – mercio
    Feb 17, 2016 at 23:37

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