Prove that the map $c:{\mathbb {T}}^{3}=S^{1}\times S^{1}\times S^{1}\setminus\ \Delta\longrightarrow \{\pm 1\}$ is continuous.

Question

Prove that the map $c:{\mathbb {T}}^{3}=S^{1}\times S^{1}\times S^{1}\setminus \Delta\longrightarrow \{\pm 1\}$ is continuous where $\Delta$ is the diagonal $\Delta=\{(g_i,g_j,g_k)\}$ for $i=j$ or $j=k$ or $i=k$ and $c$ has the following properties:

1)($\textit{Left-Invarianc}$): $c(ag_1,ag_2,ag_3)=c(g_1,g_2,g_3)$ for all $a,g_1,g_2,g_3\in S^1$ with $g_i\ne g_j$ for any $i$ and $j$

2)($\textit{Co-cycle condition}$): $c(g_1,g_2,g_3)-c(g_1,g_2,g_4)+c(g_1,g_3,g_4)-c(g_2,g_3,g_4)=0$ for all $g_1,g_2,g_3,g_4\in S^1$ with $g_i\ne g_j$ for any $i$ and $j$

These two conditions basically tells us that there is a certain order on circle which is left-invariant.

I want to prove or disprove the continuity of $c$ where $\{\pm 1\}$ is given discrete topology. We see that $\mathbb{T}^{3}=S^{1}\times S^{1}\times S^{1}$ is a three dimensional torus and not much of its topological facts are given in the wiki article. Please suggest where should I start.

EDIT: Image after Paul Plummer's comment

Answer 1

You can choose such a $c$ to be continuous. Note that means that the 3-torus without the fat diagonal is at least two connected components. How can we see that?

First, instead of trying to imagine a 3-torus, imagine three, possibly with multiplicity, marked points (which are colored/distinguished from each other) on the circle $S^1$ which will correspond to a point in $\mathbb T^3$. So if you "visually" see only two or one marked point that means that some marked points are overlapping (two coordinates are equal) and so is a point in the fat diagonal $\Delta$. Lets call the marked points $r,b,g$ and placed at $-1,i,1$ in $S^1$. It should be clear that you can not move the marked points so that $r,b,g$ are $-1,-i,1$ without crossing the fat diagonal.

It should also be clear, with this interpretation, that the components are fixed by the $S^1$ action, so $c$ can assigns the connected components a single value hence the function is continuous.