1
$\begingroup$

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 $g \circ f \simeq id_{\mathbb{R}^{n}}$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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"? $\endgroup$ – vanderlylic Mar 21 '16 at 22:23
  • 1
    $\begingroup$ @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. $\endgroup$ – Najib Idrissi Mar 22 '16 at 6:43
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.