I want to answer this question:

What is the homotopy cofibre of the unique map $S^{1} \rightarrow * $ ? describe the homotopy cofibre of $ X \rightarrow * $ in general.

My attempt:

I got a hint that I should find the cofibration of the map $S^{1} \rightarrow * .$ but I looked at the definition of "The Cofiber of a Map" in the book "Modern Classical Homotopy Theory" by Jeffery Strom, which is given below:

enter image description here enter image description here

But still I do not understand how to find the cofibration of my given map, could anyone help me in this please?

Also, I looked at the word "homotopy cofibre" definition in AT and "Modern Classical Homotopy Theory" by Jeffery Strom but I did not find it, could anyone tell me under which title can I find this word? or specifically at which page in either of the 2 books?

Also, I got a hint of solving this question by forming the weak homotopy pushout square and the strong homotopy pushout square but I do not know the relation of those to homotopy cofibration.

Also, I was given a hint of those diagrams:

enter image description here enter image description here

1-Actually, for the first diagram, which is a pushout diagram, I do not understand why $X \bigsqcup {*} = X$? does this because $X$ is a pointed space?

2-For the second diagram, I do not understand which side of the given pushout square represents our given map $S^{1} \rightarrow * ,$ is it the upper side or the left side? and why we should construct a diagram containing 2 "*"?

Could anyone help me answer this questions please? I want to arrange my thoughts to conclude the solution.

  • $\begingroup$ I don't know how to answer this right away, but there are things that sound unclear. The space $\ast$ (topological space with one point) is the terminal object in the category of topological space (i.e. there is a unique map $X\to\ast$, for all space $X$). This should already answer your point "2-", both the vertical and the horizontal map are the same, by uniqueness. Also, this is a very generic fact that taking coproduct with terminal object gives an isomoprhic object, which is why $X\sqcup\ast = X$ : you can show it by proving that $X$ satisfies the universal property defining $X\sqcup\ast$ $\endgroup$ – Thibaut Benjamin Jan 30 '20 at 15:40
  • $\begingroup$ Reading what is written in the image and taking $C_f$ with the map $X\overset f\to C_f$ to be the co-fiber and not thinking about this in any other way yields that $C_f = S^{n+1}$ and the map is $S^n\to S^{n+1}$ that collapses everything to a point. $\endgroup$ – s.harp Jan 30 '20 at 15:56
  • 1
    $\begingroup$ @ThibautBenjamin No, it's taking the coproduct with the initial object that gives an isomorphic object. $2$ is not isomorphic to $1$...Luckily $*$ is also initial in the pointed category. $\endgroup$ – Kevin Arlin Jan 30 '20 at 23:13
  • $\begingroup$ Oups, my bad, I went a bit too fast on it $\endgroup$ – Thibaut Benjamin Jan 31 '20 at 9:25

Question 1: The pushout can in fact be constructed as a quotient of the disjoint sum. Of course we have $X \sqcup * \ne X$, but we indentify $a = i(a) \in X$ with $p(a) = *$ for all $a \in A$, thus we obtain $X/A$.

Question 2: The homotopy cofiber is not obtained as the pushot of your diagram. Let us more generally consider a map $f : X \to Y$. In general it is no cofibration, but the inclusion $j : X \to M_f$ embedding $X$ as the top of the mapping cylinder is one and we have $r \circ j = f$, where $r : M_f \to Y$ is the canonical strong deformation retraction. The homotopy cofiber of $f$ is then defined as the pushout

$\require{AMScd}$ \begin{CD} X @>{j}>> M_f \\ @V{p}VV @V{p'}VV \\ * @>{f}>> C_f \end{CD}

Note that if $f$ is a cofibration, then one easily show that $C_f$ is homotopy equivalent to $X/A$.

In your case $f : S^1 \to *$ we get $M_f \approx D^2$ and $C_f \approx D^2/S^1 \approx S^2$.

Edited: As Jason DeVito pointed out in his comment, for any $f : X \to *$ we have $(M_f,X) \approx (CX,X)$, thus $C_f \approx CX/X \approx \Sigma X$.

  • $\begingroup$ And it seems in general that for $X\rightarrow \ast$, $C_f = \Sigma X$. $\endgroup$ – Jason DeVito Jan 30 '20 at 16:22
  • $\begingroup$ @Jason You are right, I shall edit my answer. $\endgroup$ – Paul Frost Jan 30 '20 at 16:52
  • $\begingroup$ what is the meaning of this symbol $\approx$? $\endgroup$ – Emptymind Feb 2 '20 at 2:04
  • $\begingroup$ It means "homeomorphic". $\endgroup$ – Paul Frost Feb 2 '20 at 9:23

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