Proving Homotopy of Maps with Examples

[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.

 

[rasmhop]

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

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:

using the fact that any two maps f,g: X \rightarrow Y where Y is a convex subset of \mathbb{R}^n are homotopic

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.

 

[latentcorpse]

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

 

[rasmhop]

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

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.

 

[latentcorpse]

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.

 

[rasmhop]

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.

 

[latentcorpse]

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.

 

[rasmhop]

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

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).

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.

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.

 

[latentcorpse]

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.

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?

 

[rasmhop]

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

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

\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.

 

[latentcorpse]

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.

 

[rasmhop]

why is it important that \sigma is a homeomorphism?

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.

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.

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)

 

[latentcorpse]

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

(i) in your very last line shouldn't it read
(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?

 

[rasmhop]

(i) in your very last line shouldn't it read
(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?

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.

(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?

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.

(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?

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.