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?

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    $\begingroup$ Removing backtracks doesn’t change the homotopy type, but deleting relators does. $\endgroup$ – HJRW Apr 5 '19 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.

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    $\begingroup$ Of course! Thank you for such a clear explanation! I'm a big fan of the good stacking =) $\endgroup$ – lunchmeat Apr 6 '19 at 20:40

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