Let $K$ and $L$ be simplicial complexes such that 1) $L\subseteq K$; 2) $K$ is homotopic to $S^4$; 3) $L$ is homotopic to $S^6$.
Is the inclusion map from $L$ to $K$ null-homotopic?
Thanks!
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9$\begingroup$ The answer is not necessarily. The point is that $\pi_6(S^4)\cong\mathbb Z_2$ and the homotopically nontrivial map $S^6\to S^4$ can be homotoped to an inclusion after replacing $S^4$ with $S^4\times D^n$ for sufficiently large $n$. $\endgroup$– Igor BelegradekCommented Mar 23, 2023 at 14:56
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$\begingroup$ Thank you for the answer. Is there any result on the minimal $n$ such that the inclusion map from $S^6$ to $S^4\times D^n$ to be nontrivial? $\endgroup$– Power of TopologyCommented Mar 24, 2023 at 14:56
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1$\begingroup$ I am not sure I understand the question; my quess is that you are asking for a minimal $n$ such that any continuous map $S^6\to S^4\times D^n$ is homotopic to a PL embedding. I don't know the minimal one, but $n=6$ is enough according to Irwin's "Embeddings of polyhedral manifolds" jstor.org/stable/1970560. For $n\ge 9$ the result is true by general position. $\endgroup$– Igor BelegradekCommented Mar 24, 2023 at 15:12
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1$\begingroup$ Thanks! This is very helpful. Sorry about being unclear. Basically, I was asking for a sufficient condition that the inclusion map from $L$ to $ K$ is null-homotopic given $L\simeq S^6$ and $K\simeq S^4$. Or any conjecture? I hope this makes sense. $\endgroup$– Power of TopologyCommented Mar 24, 2023 at 15:22
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$\begingroup$ I think, you can consider the simplcial analog of mapping cylinder of the homotopically nontrivial map $S^6\to S^4$. See e.g. Hatcher's "Algebraic topology", p.183 (the proof of theorem 2C.5 on simplicial approximation of CW complexes). It will be homotopy equivalent to $S^4$ and will contain a simplicial complex PL homeomorphic to $S^6$, and I think it will have dimension $7$ (check this). $\endgroup$– Igor BelegradekCommented Mar 24, 2023 at 16:40
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