# Homotopy homework help

1. Jan 23, 2010

### latentcorpse

Prove every map $e:X \rightarrow \mathbb{R}^n$ is homotopic to a constant map.

well i said that the constant map is $c:X \rightarrow \mathbb{R}^n;x \mapsto c$
since $\{ c \} \subseteq \mathbb{R}^n$ is a clealry a convex subspace and $e(X)=\mathbb{R}^n$ is a convex subspace of $\mathbb{R}^n$, e and c must be homotopic (using the fact that any two maps $f,g: X \rightarrow Y$ where Y is a convex subset of $\mathbb{R}^n$ are homotopic).

however, i'm not sure if i can assume $e(X) \subseteq \mathbb{R}^n$ is a convex subset. probably not. any ideas?

thanks.

2. Jan 23, 2010

### rasmhop

Re: homotopy

You can't assume that, but you don't need to. The usual straight-path homotopy:
$$H(s,t) = (1-t)e(s) + tc$$
still works since $\mathbb{R}^n$ is convex. You seem to be confusing the image and codomain. e has codomain $\mathbb{R}^n$, but its image e(X) is just a subset of $\mathbb{R}^n$. You wrote:
If you know this, then this is actually sufficient since in the case of e we have that Y is $\mathbb{R}^n$ and this is a convex subset of itself. In this quote e(X) is never mentioned and it doesn't matter that e isn't surjective.

3. Jan 23, 2010

### latentcorpse

Re: homotopy

so it's as easy as saying that since $Y=\mathbb{R}^n$ for both the map e and the map c and $\mathbb{R}^n$ is a convex subset of itself, e and c both map to convex subsets of $\mathbb{R}^n$ and are therefore homotopic by the homotopy $H: X \times I \rightarrow \mathbb{R}^n ; (s,t) \mapsto (1-t)e(s)+tc$

do i need to also show that this is in fact a homotopy?
i.e. show $h(s,0)=e(s),h(s,1)=c(s)=c$?
or is the above enough

thanks

4. Jan 23, 2010

### rasmhop

Re: homotopy

Well that depends on how much detail you want to give. If you want to do it in great detail you would have to argue that h is continuous since it's the composition of continuous functions, and that it's in fact a homotopy from e to c. In my opinion both statement are so simple that I wouldn't bother to give more detail than you gave since you stated that h is from e to c and it's very simply to verify h(s,0) = (1-0)e(s)+0c=e(s) and h(s,1) = (1-1)e(s)+ 1c=c.

5. Jan 24, 2010

### latentcorpse

Re: homotopy

ok. thanks.

the next bit asks me to show that if $f:X \rightarrow S^n$ is not surjective then f is homotopic to a constant map.

well this means image is not codomain so f maps to f(X) but i can't find any useful theorems in my notes to prove there exists a homotopy. any suggestions?

thanks.

6. Jan 24, 2010

### rasmhop

Re: homotopy

We can find some $N \in \mathbb{S}^n$ such that N isn't in the image of f, so we can restrict its codomain to form a continuous map $f' : X \to \mathbb{S}^n \setminus \{N\}$ such that $f = \iota \circ f'$ where $\iota : \mathbb{S}^n\setminus\{N\} \to \mathbb{S}^n$ is the inclusion map.

If $N \in \mathbb{S}^n$, then $\mathbb{S}^n \setminus \{N\}$ is homeomorphic to $\mathbb{R}^{n}$ by the stereographic projection $\sigma : \mathbb{S}^n \setminus \{N\} \to \mathbb{R}^{n}$. You can use this fact to construct a homotopy from $\sigma \circ f' : X \to \mathbb{R}^n$ to a constant map, and you can then use the inverse stereographic projection $\sigma^{-1}$ to obtain your desired homotopy.

7. Jan 24, 2010

### latentcorpse

Re: homotopy

but how do you know that there is only one such $N \in S^n$ that isn't in the image of $f$?

also, couldn't i just say that from the argument above (i.e. in the earlier posts) that $\sigma \circ f' : X \rightarrow \mathbb{R}^n$ is homotopic to a constant map as it is an example of such a function $e: X \rightarrow \mathbb{R}^n$.

8. Jan 24, 2010

### rasmhop

Re: homotopy

I don't. But I don't need to. We just need to remove point to get the homeomorphism, and we never state that f' is surjective (because we don't need that).

Yes that was what I had in mind. Then you get a homotopy $h : X\times I \to \mathbb{R}^n$ You then compose it with $\iota \circ \sigma^{-1} : \mathbb{R}^n \to \mathbb{S}^n$ and show that $\iota \circ \sigma^{-1} \circ h$ is your desired homotopy.

9. Jan 24, 2010

### latentcorpse

Re: homotopy

so surely i can just write: and from part a) this works rather than ahving to write out the explicit homotopy.

also how can the function $\iota$ exist? wouldn't it have to be one-to-many?

10. Jan 24, 2010

### rasmhop

Re: homotopy

For the existence of h yes. But this still only shows them homotopic in $\mathbb{R}^n$ so you need to apply $\iota \circ \sigma^{-1}$ to get back to a homotopy in $\mathbb{S}^n$.

$\iota : \mathbb{S}^n\setminus\{N\} \to \mathbb{S}^n$ is just the inclusion function, so it's simply the identity map of S^n with its domain restricted:
$$\iota(x) = x \qquad \textrm{for }x \in \mathbb{S}^n \setminus \{N\}$$
It's not surjective, but we don't need it to be.

11. Jan 24, 2010

### latentcorpse

Re: homotopy

why is it important that $\sigma$ is a homeomorphism?

also, we know h is a homotopy from the first part of the question.
how do we go about showing that $\iot \circ \sigma^{-1} \circ h$ is also a homotopy? i can't find anything that says composition of functions with a homotopy leaves a homotopy. is that where the homeomorphism plays a role? even so, $\iota$ is just a function, not a homeomorphism.

12. Jan 24, 2010

### rasmhop

Re: homotopy

Because a homeomorphism is precisely a continuous map with a continuous inverse. Thus it's exactly what is required for $\sigma^{-1}$ to exist and be continuous.

A homotopy of maps from X to Y is simply a continuous map from X x I to Y. Since $\sigma$ is a homeomorphism $\sigma^{-1}$ is continuous. $\iota = \textrm{id}|(\mathbb{S}^n \setminus\{N\})$, so $\iota$ is the restriction of a continuous function and therefore $\iota$ is continuous. Since all three functions are continuous you know that $\iota \circ \sigma^{-1} \circ h$ is continuous and therefore a homotopy. You now only need to check that it's actually a homotopy from f to some constant map.
$$(\iota \circ \sigma^{-1} \circ h)(s,0) = (\iota \circ \sigma^{-1})(h(s,0)) = (\iota \circ \sigma^{-1})((\sigma \circ f')(s)) = (\iota \circ \sigma^{-1} \circ \sigma \circ f')(s) = (\iota \circ f')(s) = f(s)$$
$$(\iota \circ \sigma^{-1} \circ h)(s,1) = (\iota \circ \sigma^{-1})(c(s)) =(\iota \circ \sigma^{-1})(c(0)) =(\iota \circ \sigma^{-1} \circ c)(0)$$

13. Jan 25, 2010

### latentcorpse

Re: homotopy

thanks but i'm still not sure about a couple of things:

$(i \circ \sigma^{-1} \circ h)(s,1)=(i \circ \sigma^{-1})(c(s))$
but then $c(s)=c$ and we don't know what $(i \circ \sigma^{-1})(c)$ is so we just leave it as $(i \circ \sigma^{-1})(c)$. now the inverse stereographic projection just maps to some point in $S^n - \{ N \}$ and $\iota$ is just the identity on this space so we remain at this point $\sigma^{-1}(c) \in S^n - \{ N \}$. since $c$ is constant, $\sigma^{-1}(c)$ will be constant and so we know we have a homotopy between f and the constant map. is that correct?

(ii) to do with the reasoning:
i accept that $\iota, \sigma^{-1}$ and $h$ are all continuous and that's grand. but only one of them is a homotopy so how can we deduce that the composition is a homotopy? surely this conclusion should come at the very end of your argument once we've verified the composition is a continuous map from f to a constant map (as this is the definition of a homotopy from f to c). is it not?

(iii) i thought all homotopies mapped from something crossed with I (the unit interval). this composition of maps just maps from some space $X$ not $X \times I$ does it not?

14. Jan 25, 2010

### rasmhop

Re: homotopy

This is correct and essentially the same as mine. Since c is constant c(0) in my post is the same as what you call c.

A homotopy in general is nothing but a continuous map from I x X to Y for some space X,Y. We don't really know what this is a homotopy between, but if we have a map from I x X to S^n, then all we really need to check is that it's continuous. Thus it suffices to check that the functions are continuous. At this point we know that it's a homotopy, but we haven't yet verified what it's a homotopy from and to so that's what the last calculation is for.

The composition is:
$$X \times I \xrightarrow{h} \mathbb{R}^n \xrightarrow{\sigma^{-1}} \mathbb{S}^n \setminus \{N\} \xrightarrow{\iota} \mathbb{S}^n$$
so it's a map from X x I to S^n.

Maybe the argument is easier to see in general. Let $H : X \times I \to Y$ be some homotopy from $p : X \to Y$ to $q : X \to Y$, and let $g : Y \to Z$ be a continuous map. Then $g \circ H : X \times I \to Z$ is a continuous map and therefore a homotopy from g(H(s,0)) = g(p(s)) to g(H(s,1))=g(q(s)). Thus we get a homotopy $g \circ H : g \circ p \simeq g \circ q$. This establishes the following result:
Thm: Let $p,q : X \to Y$ and $g : Y \to Z$ be continuous maps. Then $p \simeq q$ imply $g \circ p \simeq g \circ q$.