3
$\begingroup$

I wrote a proof that any path connected space is contractible which is completely wrong but i was not able to see what goes wrong in my proof:

Let $X$ be a path connected space. Let $P$ be point in $X$. Write $r$ the retraction $$r:X\rightarrow P;\;\; x\mapsto P$$ For any $x\in X$, take a path $$\gamma_{p,x}:I\rightarrow X$$ that starts in $P$ and ends in $x$. Now consider the homotopy $$H_t:X\rightarrow X;\;\; x\mapsto \gamma_{p,x}(t)$$ Obviously we have $H_0=i\circ r$ where $i:P\rightarrow X$ the inclusion map and $H_1=id_X$. Hence $i\circ r$ is homotopic to $id_X$ which means that $X$ deformation retracts onto $P$ and in particular that $X$ is contractible.

Now that i'm re-reading what i have written i think i see that the problem is that the choice of the path is not made in a continuous way, because there may be different paths starting in $P$ and ending in $x$, is this the problem? and if yes does that mean that if we are able to choose such a path in a canonical way then our space is contractible, for example the Cone on $X$ is contractible because we can choose canonically the path as the line segment between any point of the cone to the apex? thanks for your help!!

$\endgroup$
  • 4
    $\begingroup$ Well yes, it's the problem of course. Choosing a path "continuously" is exactly the same thing as finding a homotopy between the identity and the constant map. Think about it... $\endgroup$ – Najib Idrissi Nov 10 '15 at 11:06
  • 7
    $\begingroup$ In general when you know a proof is wrong, a standard strategy for "debugging" is to pick the simplest counterexample you know of and go through the proof step by step to see what steps fail. Here the simplest counterexample is (arguably) the circle.... :) $\endgroup$ – Andrew D. Hwang Nov 10 '15 at 11:33
  • 1
    $\begingroup$ If we look at it from a broader perspective I think that the infinite real projective space is the simplest counterexample since it has the smallest obstruction to be non contractible since its an Eilenberg-Mclane-Space for a smaller group. $\endgroup$ – ThorbenK May 20 '16 at 12:34

Your Answer

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

Browse other questions tagged or ask your own question.