The fundamental class of diagonal of torus I have the following question regarding the class of diagonal of torus $S^1 \times S^1$ - it is equal to the sum of fundamental classes of $\{pt\} \times S^1$ and $S^1 \times \{pt\}$. Can someone explain to me why that is? Here the diagonal is $\Delta \subset S^1 \times S^1$, where $\Delta = \{(x, x) \in S^1 \times S^1; x \in S^1\}$.
 A: For a space $X$ the fundamental class of itsdiagonal is the image of its fundamental class $[X]\in H_n(X)$ under $\Delta_*:H_n(X)\rightarrow H_n(X\times X)$. From the Kunneth theorem this class is of the form
$\Delta_*[X]=x\times 1+1\times y+ other$
and using the fact that $pr_i\circ \Delta=id_X$ we find that $x=y=[X]$. (We also have that $T\circ \Delta=T$ where $T$ is the twist map which interchanges the factors of the product, so $x=y$ for this reason, and all 'other' classes will be symmetric.)
Now $X=S^1$ is torsion free so the Kunneth map is an isomorphism $-\times-:(H_1(S^1)\otimes H_0(S^1))\oplus (H_0(S^1)\otimes H_1(S^1))\xrightarrow{\cong} H_1(S^1\times S^1)$. This means that there can't be any 'other' classes in $\Delta_*[S^1]$ so we get 
$[S^1\times S^1]=\Delta_*[S^1]=[S^1]\times 1+1\times [S^1]$
You could also use the multiplication $m:S^1\times S^1\rightarrow S^1$ to see this. You have $x+y=m_*[S^1\times S^1]=m_*\Delta_*[S^1]=2[S^1]$ and use the fact that $H_1(S^1)$ is free to conclude the result.
