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

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?

• Removing backtracks doesn’t change the homotopy type, but deleting relators does. – HJRW Apr 5 '19 at 5:44

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.