Show that two different embeddings of the figure-eight in the torus are not homotopic Note, we can express the torus $|X.| \cong T$ as a square with edges denoted by $e$ and $f$, the diagonal by $g$, and faces $T_1$ and $T_2$, and a single vertex $v$, with appropriate identifications.
Let $Y.$ be the semisimplicial set where $Y_1 = \{a, b\}$, $Y_0 = \{u\}$, and $Y_n = \emptyset$ for all $n \geq 2$. We can consider two maps as follows:
$\phi: Y. \rightarrow X.$ defined by $a \mapsto e$, $b \mapsto f$.
$\psi: Y. \rightarrow X.$ defined by $a \mapsto e$, $b \mapsto g$. 
How does one show that these two embeddings are not homotopic? 
(I've drawn them out, and the result seems obvious, but I don't know how to prove it. Basically, the first one is the intersection of a longitudinal loop with a latitudinal loop. The second embedding has a loop going around the inside and out, and one kind of diagonal loop composed of both of the non-trivial loops on the torus.) 
I posted this as part of a larger question here but received little help: Playing with the torus and semisimplicial sets (prove that $\phi$ and $\psi$ are not homotopic)

First, show that $H_1(T)$ can be viewed as the quotient $\mathbb Z <e,f,g>$ by the relation $g = e+f$, then observe that it is freely generated by $e,f$. Now, we see that $H_1(Y)$ is freely generated by $a,b$. Now, in terms of the basis $e,f$, the first map is $a\mapsto e, b\mapsto f$, while the second map is $a\mapsto e, b\mapsto e+f$. If these two maps were homotopic, their induced maps on homology would be equal, but they're not.
