Show that, for any space X, [X,I] has one element.

Question

My attempt: a function $h(x,t) = tf(x) + (1-t)g(x)$ (i.e. the straight-line homotopy) would work with any $f$ and $g$.

Is this too simple? It seems like the straight-line homotopy works in a lot of situations. Are there anything special about the spaces X and I in this case that makes the straight-line homotopy works?

Answer 1

You are correct, but just for the sake of completeness and "being sure of oneself", I've tried to dissect this.

Given two continuous functions $f, g : X \longrightarrow I$, you want to show that there is some continuous function $H: X \times I \longrightarrow I$ with $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$. This would show that any two continuous functions mapping $X$ to $I$ are homotopic, and hence the set of $\mathcal C(X,I)$ of continuous functions from $X$ to $I$ has a single homotopy type $($ie, $[X, I]$ has a single element$)$.

Your proposal $H(x,t) = (1-t)\,f(x) + t\,g(x)$ clearly satisfies $H(x,0) = f(x)$ and $H(x,1) = g(x)$. Moreover, for any $(x,t)\in X\times I$ we have

$$0 \leq \min \{f(x), g(x)\} \leq H(x, t) \leq \max \{f(x), g(x)\} \leq 1$$

so $H(x,t) \in I$. It follows that $H: X \times I \longrightarrow I$ as required, and it's easy to check from the formula that it is continuous, which completes the proof.

Answer 2

Your solution is correct. The underlying property of $I$ that makes it so that $[X, I]$ has one element is that $I$ is contractible, meaning that it is homotopy equivalent to the one-point space.

Indeed, the inclusion $\{ 0 \} \hookrightarrow I$ and the unique map $I \to \{ 0 \}$ are homotopy-inverses, with the composite $I \to \{ 0 \} \hookrightarrow I$ being homotopic to $\mathrm{id}_I$ via linear homotopy. This is why $f \sim g$ via linear homotopy for all continuous maps $f,g : X \to I$.

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More generally, suppose $C$ is a contractible space, so that $C \simeq \{ \star \}$. Let $u : C \to \{ \star \}$ be the unique map and let $v : \{ \star \} \to C$ be its homotopy-inverse, so that $v \circ u : C \to C$ is homotopic to $\mathrm{id}_C$ via some homotopy $H : C \times I \to C$.

Given any two continuous maps $f,g : X \to C$, we have $v \circ u \circ f = v \circ u \circ g : X \to C$, since these maps both factor through the one-point space (and hence are constant, with the same value).

Now you can use the homotopy $H$ to construct a homotopy $f \sim g$ by noting that $$f = \mathrm{id}_C \circ f \sim v \circ u \circ f = v \circ u \circ g \sim \mathrm{id}_C \circ g = g$$ Explicitly, the homotopy $K : f \sim g$ looks something like $$K(x,t) = \begin{cases} H(f(x),2t) & \text{if } 0 \leq t \leq \frac{1}{2} \\ H(g(x),2(1-t)) & \text{if } \frac{1}{2} \leq t \leq 1 \end{cases}$$

It follows that $[X,C]$ has one element for all contractible spaces $X$.