By "backtrack" I mean a subword of a relator in a group presentation of the form $x x^{-1}$.

Let $X = \langle a \rangle$ as a presentation complex.

Let $Y = \langle a$ | $aa^{-1} \rangle$ as a presentation complex.

Now we see that $X$ is a circle and $Y$ is a pinched torus, and these two spaces clearly do not have the same Homotopy Type as $\pi_2(Y)$ is nontrivial.

However it was said in "A Covering Space With no Compact Core" (MSN) by Daniel Wise that:

$\langle a, b, t $ | $ [a,b]^t = [a,b][b,a] \rangle$ is homotopy equivalent to $\langle a, b, t $ | $[a,b] \rangle$.

Is this always true when the backtrack is a proper subword of a relator? Is the above case with $X$ and $Y$ the only real nonexample?

  • 1
    $\begingroup$ Removing backtracks doesn’t change the homotopy type, but deleting relators does. $\endgroup$ – HJRW Apr 5 at 5:44

Let me elaborate on my comment a little. It's a standard exercise that attaching a cell by homotopic maps leads to homotopy-equivalent spaces. (See p.13 of Hatcher, for instance.) In particular, it certainly is true that deleting backtracks leads to a homotopy equivalence of presentation complexes.

From this we conclude that your complex for $Y$ is homotopy equivalent to the presentation complex of $Z=\langle a\mid 1\rangle$ which is, of course, $S^1\vee S^2$. In particular, $\pi_2(Y)\cong\pi_2(Z)$ is non-zero, as you noted.

However, your first presentation $X$ is obtained from $Z$ by deleting a (trivial) relator, which corresponds to deleting the 2-cell in the presentation complex of $Z$. Of course, deleting a 2-cell does change the homotopy type.

In particular, Wise's assertion is correct. If you want, the above might be taken to justify your hope that $X$ and $Y$ are the only "real" counterexamples.

  • 1
    $\begingroup$ Of course! Thank you for such a clear explanation! I'm a big fan of the good stacking =) $\endgroup$ – lunchmeat Apr 6 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.