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Let $F$ be the homotopy fiber of the inclusion $X \rightarrow X \times X.$

(1)Show that $\pi_{i}(F) \cong \pi_{i +1}(X).$

Could anyone explain to me how to prove this, please?

N. Owad
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Intuition
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1 Answers1

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We let $f: E \rightarrow B$ be the map obtained by converting $X \rightarrow X \times X$ into a fibration.

Explicitly we let $E = X^I$, $B = X \times X$ and if $p \in X^I$ is a path $I \rightarrow X$, $f(p) = (p(0),p(1))$. The important property that $f$ has is that it is a fibration and is pointwise homotopy equivalent to $X \rightarrow X \times X$.

Let $x_0 \in X$ be the base point.

The fiber of $f$ is $\Omega X$ since if we let $(x_0, x_0)$ be the base point of $X \times X$, $f^{-1}((x_0, x_0)) = \{p:I \rightarrow X: (p(0),p(1)) = (x_0, x_0)\}$, the set of all paths in $X$ with start and end point equal to the base point.

Thus since $f$ is a fibrant replacement for $X \rightarrow X \times X$, we know that $F \simeq f^{-1}(x_0,x_0) = \Omega X$ and this isomorphism you are looking for reduces to proving that $\pi_i(\Omega X) = \pi_{i+1}(X)$ which you can find proofs for if you search for "suspension-loop adjunction".

Noel Lundström
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  • Can you give a reference for a good book for this proof, please? – Intuition Mar 20 '20 at 14:35
  • I recommend "Modern classical homotopy theory" which I see that you have been asking questions about. The chapter on homotopy limits will contain what you need. The homotopy fiber of a map $f:A \rightarrow B$ is the homotopy limit of the diagram $* \rightarrow B \xleftarrow f A $ and if $\mu : I \rightarrow J$ is a natural transformation between diagrams and is a pointwise homotopy equivalence then $\mu$ will induce homotopy equivalence of the homotopy limit. And the homotopy fiber of a fibration is just the fiber at the base point. With these facts you should be able to prove this yourself. – Noel Lundström Mar 20 '20 at 20:21