My bad. In my mind, $K\A$ is the induced subcomplex from $K$ with the vertex set $K^{(0)}\setminus A^{(0)}$. Thanks a lot!

What is an iterated cone pls? I think $K$ may not be contractible if $A$ is not. For example $K$ is the surface of square bipyramid and consider $A$ to be square; this is a suspension of a square, i.e. $\simeq S^2$, also a join of $S^0$ and $S^1$.

Yes. I didn't recognize that! thanks a lot.

Thanks a lot. I will thick that proof out.

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.

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?