# Playing with the torus and semisimplicial sets (prove that $\phi$ and $\psi$ are not homotopic)

Recall that we can express the torus $|X.| \cong T$ as a square with edges $e$ and $f$, diagonal $g$, faces $T_1$ and $T_2$, and a single vertex $v$, and 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 $a \mapsto e$, $b \mapsto g$.

Which map from the "figure-eight" to the torus. The first map seems to take the figure-eight to the traditional two non-trivial loops (going inside the hole then back out, and going around the equator). The second map, on the other hand, seems to take the first loop to the same loop on the torus, but the second loop to a combination of the first and second (a kind of diagonal loop composed of one of each of the non-trivial loops from before).

My goal is to compute the kernel and cokernel of $\phi_*$ and $\psi_*$ for each of $H_0, H_1, H_2$, and also to prove that $\psi$ and $\phi$ are not homotopic (which seems intuitively obvious, but for which the proof is escaping me).

• Could you let me know what you are using for a reference for this material? For example, have you learned a way to compute the induced map on homology from a simplicial map? If you have that, you can compute $\phi_*$ and $\psi_*$. – Ryan Budney Sep 20 '13 at 0:40
• I'm familiar with these kinds of arguments, but it'd be helpful to be walked through it again (it's been a long time). I've thought about it some more, and I'm pretty sure that $\phi_*$ is surjective but not injective... – user95780 Sep 20 '13 at 0:42
• Could you let me know what reference you are using for your definitions? – Ryan Budney Sep 20 '13 at 0:44
• Hatcher, for the most part. – user95780 Sep 20 '13 at 0:45
• Oh, okay. Computing the kernel and cokernel (at least, the isomorphism types) will not suffice to answer the latter part of your question but computing the image of the map on $H_1$ does. – Ryan Budney Sep 20 '13 at 0:47

For the figure eight $\large\infty$ the chain complex is $$0\to \langle a,b \rangle\to \langle u \rangle\to 0$$ where all maps are $0$, so the these cycles also generate the homology groups $$H^Δ_2=0,\quad H^Δ_1=⟨a,b⟩=\Bbb Z^2, \quad H^Δ_0=⟨u⟩=\Bbb Z$$
For the torus $T^2$ the chain complex is $$0 →⟨T_1,T_2⟩ →⟨e,f,g⟩ →⟨v⟩ →0$$ where only the first map is non-trivial and sends $T_i$ to $e+f-g$. The simplicial homology groups are $$H^\Delta_2= ⟨T_1-T_2⟩=\Bbb Z,\quad H^Δ_1=\frac{⟨e,f,e+f-g⟩}{⟨e+f-g⟩}=⟨e,f⟩=\Bbb Z^2, \quad H^Δ_0=\Bbb Z$$ To show that $\phi_*,\psi_*$ are not homotopic, consider the commutative square $$\begin{array}{ccc} H^\Delta_n(\infty) & \to & H_n(\infty)\\ \downarrow & & \downarrow\\ H^\Delta_n(T^2) & \to & H_n(T^2) \end{array}$$ The horizontal maps are isomorphisms between the simplicial and the singular homology groups, and these morphisms respect the maps $\phi_*$ and $\psi_*$. But homotopic maps induce the same maps on singular homology groups, thus also on simplicial groups. However, $\phi_*$ and $\psi_*$ are not equal. The first sends $a$ to $[e]$ and $b$ to $[f]$, and is an isomorphisms. The second one sends $a$ to $[e]$ and $b$ to $[g]$ which is equal to $[e+f]=[e]+[f]$, and is an isomorphisms, too, but a different one.