Coproduct of the homology coalgebra of the sphere Let $S^m$ be the $m$-sphere and $H_*(S^m)$ be the homology coalgebra with field coefficient. Then what is the coproduct of
$
H_*(S^m)
$?
For $x$ the generator of $H_*(S^m)$, does
$$
\Delta_*x=0?
$$
or 
$$
\Delta_*x=1\otimes x+x\otimes 1?
$$
How to compute?
 A: With field coefficients $\Bbbk$ and a finite type space $X$ (i.e. all the homology groups are finite-dimensional), the homology coalgebra $H_*(X)$ is dual to the cohomology algebra $H^*(X)$ equipped with the cup product, see e.g. this answer by Ben Webster on MO. I'll try to write down all the details.
In the case of the sphere, if $1 \in H^0(S^m) \cong \Bbbk$ and $u \in H^m(S^m) \cong \Bbbk$ are generators of both vector spaces, then it's well known that:
$$1 \smile 1 = 1, \; 1 \smile u = u \smile 1 = u, \;u \smile u = 0.$$
Suppose $x \in H_m(S^m)$ is dual to $u \in H^m(S^m)$, and denote by the same letter $1 \in H_0(S^m)$ the dual of $1 \in H^m(S^m)$. The coproduct $\Delta : H_*(S^m) \to H_*(S^m) \otimes H_*(S^m)$ is dual to the cup product $H^*(S^m) \otimes H^*(S^m) \to H^*(S^m)$ (using the Künneth isomorphism). For example $\Delta(x)$ is an element of
$$\Delta(x) \in \left( H_*(S^m) \otimes H_*(S^m) \right)_m = H_0(S^m) \otimes H_m(S^m) \oplus H_m(S^m) \otimes H_0(S^m).$$
A basis of $H_0(S^m) \otimes H_m(S^m)$ is given by $\{1 \otimes x\}$, and a basis of $H_m(S^m) \otimes H_0(S^m)$ is given by $\{x \otimes 1\}$, so there are scalar $\lambda, \mu \in \Bbbk$ such that $\Delta(x) = \lambda \cdot (1 \otimes x) + \mu (x \otimes 1)$. Since $1 \smile u = u$, it follows that
$$1 = \langle x, u \rangle = \langle x, 1 \smile u \rangle = \langle \lambda \cdot (1 \otimes x) + \mu (x \otimes 1), 1 \smile u \rangle = \lambda \cdot 1 + \mu \cdot 0 = \lambda$$ (recall that $x$ is the dual of $u$). Similarly $u \smile 1 = u$ so the coefficient $\mu$ in front of $x \otimes 1$ is $1$ too. It follows that $\Delta(x) = 1 \otimes x + x \otimes 1$. Similarly $\Delta(1) = 1 \otimes 1$.
Note that $\Delta(x) = 0$ wasn't possible. The homology coalgebra is counital: there is a counit $\epsilon : H_*(X) \to \Bbbk$ (where $\Bbbk$ is concentrated in degree zero) such that for all $x$, if $\Delta(x) = \sum x_{(1)} \otimes x_{(2)}$, then $x = \sum \epsilon(x_{(1)}) x_{(2)} = \sum \epsilon(x_{(2)}) x_{(1)}$. This isn't possible if $\Delta(x) = 0$ (unless $x = 0$ itself). The counit is given here by $\epsilon(1) = 1$, $\epsilon(x) = 0$.
