Let $(H, m, u, \Delta, \epsilon, S)$ be a Hopf algebra and $h\in H$. Do we have the equality
$$u\epsilon(h)\otimes h = h$$
Or maybe put $1\otimes h$ on the RHS or something else "isomorphic"?
EDIT: To give the context: this is what the exercise asks me to prove:
$$\sum_{(h)} h_{(1)}S(h_{(2)})\otimes h_{(3)} = h$$
My solution is: $$\sum_{(h)} h_{(1)}S(h_{(2)})\otimes h_{(3)} \\= ((m \circ(I\otimes S))\otimes I)(\Delta \otimes I) (h) \\= (m\circ (I\otimes S)\circ \Delta) \otimes I (h) \\= u\epsilon (h) \otimes h $$
Now the step I would need to make is the one in the title. Or have I already made a mistake in my calculations?
EDIT2: I see I made the mistake: $\Delta_2$ should be $(I\otimes\Delta)\Delta$ so what we are actually left with is
$$(u\epsilon \otimes I) \Delta(h)$$
But how does this simplify to $h$ (or $h\otimes 1 \otimes 1$ or something like that)?
EDIT3 Ok, now I see it:
$$(u\epsilon \otimes I) \Delta(h) = \sum_{(h)} \epsilon(h_{(1)})h_{(2)} = h$$ by the counital property as mentioned. (Or maybe you put the tensor product in there or not; there's the natural isomorphism $\mathbb{K}\otimes H \simeq H$, but I guess that goes away when you multiply: $1\otimes h = 1h = h$).
Thank you very much!
Let me try to put some order, just of the sake of clarity. The antipode condition says that $$\label{antipode}\tag{$\dagger$} \sum_{(h)}h_{(1)}S(h_{(2)}) = \varepsilon(h)1_H = \sum_{(h)}S(h_{(1)})h_{(2)} $$ for every $h\in H$, that is to say, $$\label{ant2}\tag{$\ddagger$} I*S=m\circ(I\otimes S)\circ \Delta = u \circ \varepsilon = m\circ (S\otimes I)\circ \Delta=S*I. $$ You want to prove that $$\sum_{(h)}h_{(1)}S(h_{(2)})\otimes h_{(3)} = 1_H\otimes h$$ (which seems to me the only reasonable thing to prove, by reading your post, and it is one of the exercises on Sweedler's book, whence it seems to me a reasonable exercise that could be given to a student).
Solution 1 (elementwise) \begin{align} \sum_{(h)}h_{(1)}S(h_{(2)})\otimes h_{(3)} & = \sum_{(h)}h_{(1)_{(1)}}S(h_{(1)_{(2)}})\otimes h_{(2)} \stackrel{\eqref{antipode}}{=} \sum_{(h)}\varepsilon(h_{(1)})1_H\otimes h_{(2)} = 1_H\otimes h. \end{align} Solution 2 (commutative diagrams) \begin{align} (m\otimes I)\circ(I\otimes S\otimes I)\circ (\Delta\otimes I)\circ \Delta \stackrel{\eqref{ant2}}{=} (u\otimes I)\circ (\varepsilon\otimes I)\circ\Delta = u\otimes I. \end{align}