Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's define the following:

(Def) A continuous map $f_0 : A \to B$ where $A,B$ are topological spaces is called essential if every homotopic map $f_1$ is surjective, i.e. $f_1 (A) = B$.

Let $B=S^n$ be the $n$-sphere and let $f_0 : A \to B$ be non-essential. Claim: Then there exists a homotopic map $f_1$ such that $f_1 (A) = \ast$ is a one point space.

That's a footnote in the paper I'm reading. But I don't understand the claim. If $A$ is simply-connected then I think I "see" that $f_0(A) = f_1(A) = \ast$. But the claim in the paper does not put any restrictions on $A$ so in particular, it could be disconnected. I'm a bit confused. Would someone show me how the claim is true? Thanks lots.

share|cite|improve this question
Inflate the point you miss and contract the rest of the image to the antipode. – t.b. Sep 7 '12 at 13:12
@t.b. But what if it looks like a bunch of scattered points? How do I contract that? – Rudy the Reindeer Sep 7 '12 at 13:15
Wait. What? I want to continuously retract the image to a point, right? – Rudy the Reindeer Sep 7 '12 at 13:16
I think you are confusing the idea of homotopic maps with homotopic spaces. Consider the map $f: \lbrace 0,1 \rbrace \to \mathbb{R}$ such that $f(0)=0$ and $f(1)=1$. Then we can make a homotopy $H:\lbrace 0,1\rbrace \times [0,1] \to \mathbb{R}$ defined by $H(x,t)=xt$. Then $H(x,1)=f$ and $H(x,0)=0$. Thus while the image of $f$ is disconnected. The image of $H(x,0)$ is a one point space. – Kris Williams Sep 7 '12 at 13:53
@DrKW Thank you very much for this comment. I did indeed confuse the images of homotopic maps with homotopic spaces. – Rudy the Reindeer Sep 7 '12 at 14:21
up vote 2 down vote accepted

You confused homotopic maps with homotopic spaces, as pointed out by DrKW in the comments. Here is what you can do to the image if $f$ is not surjective:

enter image description here

This does not work if $f$ is surjective since then you'd have to "tear" the image. But of course something that "tears" your space cannot be continuous.

share|cite|improve this answer
Or $f_0$, as I call it in the drawing. – Rudy the Reindeer Sep 7 '12 at 15:27
That is an excellent diagram. – Kris Williams Sep 7 '12 at 15:41
@DrKW Thank you : ) – Rudy the Reindeer Sep 7 '12 at 15:42
+1 for following the advice of drawing pictures :) – t.b. Sep 7 '12 at 15:43
@t.b. Been following that (= your advice) for a loong time now : ) – Rudy the Reindeer Sep 7 '12 at 15:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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