Compute the effect of $∆_∗$ on homology groups. 
Let $∆ : S^n → S^n × S^n$ be the diagonal map $∆(x)=(x,x)$. Compute the effect of $∆_∗$ on homology groups.

My attempt :
I considered the CW chain complex of $S^n$ and let, $H_0(S^n)=\Bbb Z\{v\}$ and $H_n(S^n)=\Bbb Z\{e\}$ . Then    $S^n × S^n$ has it's non-zero homlogy groups $H_0(S^n \times S^n)=\Bbb Z\{v \otimes v\},H_n(S^n \times S^n)=\Bbb Z\{v \otimes e,e \otimes v\},$$ H_{2n}(S^n \times S^n)=\Bbb Z\{e \otimes e\}$ .
$H_{2n}(S^n)=0$ . 
Looking at $H_0(S^n)$ , $\Delta_*  : H_0(S^n) \to H_0(S^n \times S^n)$ behaves like, $\Delta_*(v)=(v,v)$ which can we identify with $v \otimes v$ ? If it is true then it's identity on the 0th homology .
And then ,$\Delta_*  : H_n(S^n) \to H_n(S^n \times S^n), \Delta_*(e)=(e,e)$ how to understand the realtion with the generators  $v \otimes e,e \otimes v$ ?
Thanks in advance for help! 
 A: Example 3B.3 of Hatcher's Algebraic Topology can enlighten your problem a bit. I'm going to be using the Künneth formula all the time.
In the first case, as we mention, you have factorized $\Delta_*:H_0(S^n)\to H_0(S^n)\times H_0(S^n)\to H_0(S^n)\otimes H_0(S^n)\cong H_0(S^n\times S^n)$. The map between the cartesian product and the tensor product is the canonical one, sending $(x,y)\mapsto x\otimes y$. In particular, if you call $u$ the generator of the first $S^n$ of the product, $v$ the generator of the second $S^n$ of the product, and $e$ the $S^n$ of the source, it is clear by cellular homology that $e\mapsto (u,v)\mapsto u\otimes v$. You called $v$ both $u$ and $v$, which is not a problem, but I think that could have led you to a confusion. 
In the second case, we have that $H_n(S^n\times S^n)\cong (H_n(S^n)\otimes H_0(S^n))\oplus (H_0(S^n)\otimes H_n(S^n))$. Then, you can choose generators $(u\otimes v)$ and $(v\otimes u)$, (I'm now identifying generators of both spheres, you can avoid it if you want). Then, the generator $a$ of $H_n(S^n)$ is mapped $(u\otimes v)\oplus (v\otimes u)$ (if you called the generators by a different name you just have to keep the naming here). You can even call $a$ to every generator of every $H_n(S^n)$, which simplifies the notation, giving you that $a\mapsto (a\otimes v)\oplus (v\otimes a)$, which is probably more clear. Now, $(u\otimes v)$ and $(v\otimes u)$ (or $(a\otimes v)$ and $(v\otimes a)$) are just the generators of the direct sum, so everything is known. 
The important thing here, is that the generators are product of cells of dimension whose sum is the dimension of the homology.
