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](https://link.springer.com/article/10.1023%2FA%3A1019693615888)" ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=1934009)) 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? 



A Backtrack as a Single Word in a Group Presentation yields a Complex that isn't of the Same Homotopy Type?
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.


