# $X$ is contractible if and only if $X \simeq \{ * \}$ - A three-part question

First some background:

The topological spaces, $X, Y$, are homotopically equivalent if and only if there are continuous functions, $f \colon X \longrightarrow Y$ and $g \colon Y \longrightarrow X$ with $g \circ f \simeq id_{X}$ and $f \circ g \simeq id_{Y}$. Explicitly, there is homotopies $H_{1}$ and $H_{2}$ $$H_{1} \colon X \times [0,1] \longrightarrow X$$ with $H_{1}(x, 0) = g \circ f$ and $H_{1}(x, 1) = id_{X}$ and $$H_{2} \colon Y \times [0,1] \longrightarrow Y$$ with $H_{2}(x, 0) = f \circ g$ and $H_{2}(x, 1) = id_{Y}$. Write $X \simeq Y$ when this is the case.

$X$ is contractible if and only if $X \simeq \{ * \}$.

Now the questions:

(i) Show that $\mathbb{R}^{n}$ is contractible for every counting number, $n$.

(ii) Show that if $f \colon X \longrightarrow Y$ is continuous and $Y$ is contractible, then $f$ is homotopic to a constant function.

(iii) Find an example of a contractible topological space $X$ and a continuous function $f \colon X \longrightarrow Y$ which is not homotopic to a constant function.

Where I am having trouble:

I feel for all the questions I am completely lacking a starting point and an intuition about what I should be doing to prove each part. Do I need to construct explicit maps, or appeal to general properties? I know about homotopy between two functions and that it is an equivalence relation but not much else. It would be helpful to know a solid starting point of what I should use for each question.

EDIT: (i) Okay, here is some progress I made. Take $f \colon \mathbb{R}^{n} \rightarrow \{ * \}, x \mapsto *$ and $g \colon \{* \} \rightarrow \mathbb{R}^{n}, * \mapsto a$, for some fixed $a$. Then it is obvious that we have $f \circ g = id_{\{ * \}} \simeq id_{\{ * \}}$. However, I am unsure how we show that $g \circ f \simeq id_{\mathbb{R}^{n}}$.

## 2 Answers

Yes, you need to construct explicit maps. Take question (i) for example. Let $f : \mathbb{R}^n \to \{*\}$ and $g : \{*\} \to \mathbb{R}^n$ be as you describe; in fact, for simplicity, just let $g(*) = 0$. It's clear that $f \circ g = \operatorname{id}_{\{*\}}$. Now you want to prove $g \circ f \simeq \operatorname{id}_{\mathbb{R}^n}$.

What's the definition of that statement? This is the question you should be asking yourself. It means there exists some $H : \mathbb{R}^n \times [0,1] \to \mathbb{R}^n$ such that $H(x,0) = \operatorname{id}_{\mathbb{R}^n}(x) = x$ and $H(x,1) = g(f(x)) = 0$. Now there's some amount of guessing you need to do, but $$H(x,t) = (1-t)x$$ looks like a good candidate, and indeed it's a homotopy between $g \circ f$ and $\operatorname{id}_{\mathbb{R}^n}$.

Now for (ii): all you know is that $Y$ is contractible. What does that mean? Recalling what we did in question (i), it means that there's some $H : Y \times [0,1] \to Y$ such that $H(y,0) = y$ and $H(y,1) = y_0$ is some point in $Y$. Now take your $f : X \to Y$. It seems natural to try and compose it with $H$, I think. So define $G : X \times I \to Y$ by $G(x,t) = H(f(x),t)$.

What does this map satisfy? Well, $G(x,0) = H(f(x),0) = f(x)$, and $G(x,1) = H(f(x),1) = y_0$. What did we want to prove? That $f$ and some constant map were homotopic. Well, just look at the equations we just got: $G$ is exactly such a homotopy! And all we had to do was to calmly take a moment to recall the definitions of all the words in the question.

Now (iii) is more tricky, because I think the question is wrong...

If $X$ is contractible, it means there's some $H : X \times [0,1] \to X$ with $H(x,0) = x$ and $H(x,1) = x_0$ is some point (independent of $x$). Let $f : X \to Y$ be any continuous map. We're looking for a homotopy $G : X \times [0,1] \to Y$ such that $G(x,0) = f(x)$ and $G(x,1) = y_0$ is some point.

But we can just let $G(x,t) = f(H(x,t))$, and then $G(x,0) = f(H(x,0)) = f(x)$, while $G(x,1) = f(H(x,1)) = f(x_0)$ is independent of $x$ (you can call $y_0 = f(x_0)$). So $f$ is always homotopic to a constant map.

• This was very useful. For (i), would it be would it be simpler to have $H(x,t) = xt$ with $H(x,0) = 0$ and $H(x,1) = x$, given that homotopy is symmetric. Also from (ii) and (iii) could we then conclude that "$f \colon X \rightarrow Y$ is homotopic to a constant function, if and only if $X$ and $Y$ are contractible"? – vanderlylic Mar 21 '16 at 22:23
• @vanderlylic For (i) it doesn't really change anything, you can do it like that if you prefer. For (ii) and (iii), no, you cannot conclude that. A constant map $S^1 \to S^1$ is of course homotopic to a constant map, but $S^1$ is not contractible. What is true is that for a fixed $X$, if for all spaces $Y$ all maps $X \to Y$ are nullhomotopic, then $X$ is contractible. Similarly for a fixed $Y$, if for all $X$ all maps $X \to Y$ are nullhomotopic then $Y$ is contractible. – Najib Idrissi Mar 22 '16 at 6:43

You should be constructing specific things at this point, while you're still getting the basic definitions down. There's time enough later to handwave!

i) What is $g \circ f: \mathbb{R}^n \to \mathbb{R}^n$? It takes $x \mapsto a$ for all $a$. There's an "obvious" homotopy $H(x, t)$ such that $H(x, 0) = x$ and $H(x, 1) = a$; can you tell me what it is? A homotopy is basically a continuous-over-time deformation of space; can you take all of space to this one point in a continuous way?

ii) If $Y \simeq \{ * \}$ then there are continuous functions $a: Y \to \{ * \}$ and $b: \{* \} \to Y$ such that $ab \simeq 1_{\{*\}}, ba \simeq 1_Y$. That is, we have two homotopies $H_1$, $H_2$. We've already got a continuous map $f: X \to Y$, so how can we use those data to get a homotopy $H(x, t): X \times [0,1] \to \mathcal{S}$ (some space $\mathcal{S}$) between $f$ and a constant function?