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Problem of the week

Let \(G = (S | R)\) be a group presentation where S is a set of generators and R is a set of relators. Given any subset S' of S, we set R' to be the subset of R which consists of words only in the elements of S'. Then the presentation \(S'|R'\) is called a sub-presentation of \(S|R\).

The presentation complex for the presentation \(S|R\) is a cell complex constructed as follows: start with a single vertex v. For each element s of S, we attach an oriented edge labelled by s to v by identifying both endpoints of the edge with v. In this way, we get a wedge of circles where the circles are in 1-1 correspondence with the generating set S. For each element r of R, we attach a closed disk to the wedge of circles we obtained so that the boundary of the disk after gluing can be read using the labels of the edges to be the word r we started with.

For instance, consider the following presentation of a group \(x, y| xyx^{-1}y^{-1}\). We get a wedge of two circles first labelled by x and y. Then we add one disk so that the boundary reads as \(xyx^{-1}y^{-1}\). It is easy to see that the resulting space is homeomorphic to the torus. As one sees from this example, the presentation complex is a cell complex whose fundamental group is the group with the given presentation. For a group presentation \((S|R)\), let \( K(S|R) \) denote the presentation complex.

Suppose we have a group presentation \((S|R)\) such that any continuous map \(f: S^2 \to K(S|R) \) is homotopic to a constant map where \(S^2\) is the 2-sphere. Prove or find a counter-example to that every sub-presentation of \((S|R)\) has the same property, i.e., for any sub-presentation \((S|R)\), every continuous map \(h: S^2 \to K(S'|R')\) is homotopic to a constant map.