# Equivalence of two elements in $\pi_2(BX)$, where $BX$ is the rack space corresponding to the quandle $X$.

I am reading the proof of Lemma 4.1 given in the chapter "Some of Quandle Cocycle Invariants of links" of the book "Quandles and Topological Pairs" by "Nosaka." Before coming to the question, I have to define some terms.

Let $$D$$ be a link diagram over $$\mathbb{S}^2$$, and is colored by a quandle $$X$$. Consider the rack space $$BX$$, where the $$0$$-skeleton is a single vertex, $$1$$-skeleton is wedge of $$|X|$$ many circles, and the $$2$$-skeleton is constructed by attaching $$2$$-cells. Each $$2$$-cell, labelled as $$(a,b)$$ where $$a, b \in X$$, is shown below:

.

Now thicken the diagram $$D$$ so that the surface divides into three types of regions:

• Square regions about each crossing,
• Ribbon regions about the parts of arcs which don't lie in the square regions and
• Outer regions consisting of everything else.

Now define a map $$f: \mathbb{S}^2 \to BX$$ as below:

• Send each square to the corresponding $$2$$-cell as shown in the figure below:

.

• suppose ribbon is around the arc labeled $$a$$, then send it to the cell labeled as $$a$$,
• send the rest to the $$0$$-vertex in $$BX$$.

Now, after performing the second Reidemeister on the diagram $$D$$, we have a new map says $$\tilde{f}: \mathbb{S}^2 \to BX$$. It is said that these two maps $$f$$ and $$\tilde{f}$$ are the same in the group $$\pi_2 (BX)$$.

I don't know why $$f$$ and $$\tilde{f}$$ are equal in $$\pi_2(BX)$$?

Can someone explain it?