Just a remark, but in addition to second countable, you also need to assume Hausdorff. Otherwise you get examples like the line with two origins.

To add to @RyanBudney's answer, there is an invariant called Whitehead torsion that obstructs whether two homotopy equivalent complexes are actually simple homotopy equivalent (related by standard moves).

"Every manifold is homotopy equivalent to a compact one": if you want a connected counterexample, take an infinite connect-sum (which is not well-defined in the infinite case, but just choose one.) Or take the complement of a cantor set in $\mathbb R^n$ for a more sophisticated (and related) example.

The line with two origins is a simple counterexample.

@ChrisPressey I tried this out on someone who stubbornly refused to believe .99999...=1, and actually saw himself as standing up to the ivory tower mathematicians who believed infinity was real. (Ugh.) His response was: that 1-.9999...= 0.0̅1. I had to admit it was a clever, if meaningless, answer.

A story I heard is that Stephen Smale proved that sphere eversion is possible for even dimensional spheres using vary general arguments, but there was so much doubt about his result that he went ahead and constructed a specific sphere eversion for the $2$-sphere.

Does "k-fold cyclic" mean "k-fold cyclic cover?"

Simple answer: the given knot is obviously homotopic to the knot with winding number $1$ and homotopy implies homology.

Maybe I am misunderstanding the question, but could you not take $d=2,t=1$ and $A$ a rotation matrix? Then you could pretty much get whatever $m$ you want.

Doyle and Conway's division by three paper.

Indeed, one needs to identify collars rather than glue along the boundary in order to get a differentiable structure on the result.

@nombre indeed. The problem is with the application of the result, not the result itself, that and loose language around quantification.

Although a knot and its mirror image do cancel in the concordance group!