Suppose that a space $A$ homotopy dominated by a space $X$. i.e., there exist continuous maps $f:A\longrightarrow X$ and $g:X\longrightarrow A$ so that $g\circ f\simeq 1_A$. Also, let $\phi :K\longrightarrow A$ be a continuous map. Put $\phi'=f\circ \phi:K\longrightarrow X$. We know that $M_{\phi}\simeq A$ and $M_{\phi'}\simeq X$, where $M_{\phi}=\frac{K\times I \cup A}{(k,1)\sim \phi (k)}$ is the mapping cylinder of $\phi$. Also, it is a well-known fact that $\pi_1 (K)$ acts on the whole long exact sequence of homotopy groups for $(M_{\phi},K\times \{ 1\})$ and $(M_{\phi'},K\times \{ 1\})$, the action commuting with the various maps in the sequence.

My question is :

Is $\pi_2 (M_{\phi},K\times \{ 1\})$ a direct summand of $\pi_2 (M_{\phi'},K\times \{ 1\})$ as $\mathbb{Z}\pi_1 (K)$-module?


Consider $h: M_\phi \to M_{\phi’}$ which agrees with $f$ on $A$ and is the identity on $K \times I$; $h$ is well-defined by definition of $\phi’$. Then the homotopy equivalences of the mapping cylinders to their bases shows that $h$ has a left homotopy inverse, since $f$ does. Hence, $h$ induces an inclusion of a direct summand $h_\ast: \pi_2(M_\phi, K) \to \pi_2(M_{\phi’}, K)$.

This $h_\ast$ should be equivariant with respect to the action of $\pi_1(K)$, although I’m not sure how to show this at the moment.

  • $\begingroup$ I think you define $h:M_{‎\phi‎} \longrightarrow M_{‎\phi '‎}$ by $[a]\mapsto [f(a)]$ and $[k,t]\mapsto [k,t]$‎. ‎Obviously‎, ‎$h$ is a continuous map‎. ‎Also‎, ‎$r_2 \circ h=f\circ r_1$ and $h(X)=X$, where $r_1‎ :‎M_{‎\phi‎} \longrightarrow A$ and $r_2‎ :‎M_{‎\phi‎ '}\longrightarrow X$ are deformation retractions‎. ‎So‎, ‎we can consider the map $h$ as a map $(A,K)\longrightarrow (X,K)$‎. Is this correct? $\endgroup$ – M.Ramana Jun 18 '18 at 17:35
  • $\begingroup$ The pair $(A,K)$ does not make sense since we only have a map $\phi$ of $A$ into $K$. But the equation $r_2h = fr_1$ is exactly what is needed to show that $h$ has a left homotopy inverse. $\endgroup$ – Joshua Mundinger Jun 18 '18 at 17:38
  • $\begingroup$ How do I find a relative map like $d:(X,K)\longrightarrow (A,K)$ so that $d\circ h\simeq 1_{(A,K)}$? $\endgroup$ – M.Ramana Jun 18 '18 at 17:40
  • $\begingroup$ I mean the pair $(M_{\phi},K\times \{1\})$. But how? $\endgroup$ – M.Ramana Jun 18 '18 at 17:42
  • $\begingroup$ I tried to do that but there was a problem. $\endgroup$ – M.Ramana Jun 18 '18 at 17:44

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