It seems like we need this map for the coalgebra to do this problem. Or am I not understanding this correctly?
From my understanding, you need to show that if you have any such that , then has to be strictly only in , ie. is not of the form .
And you don’t think we need the counit map to show this?
What if all of the lamdas are 0 except one which is the 1 of F. Doesn’t that give us what we want?
Yes it does, but you have to derive to that conclusion via the condition of the coproduct.
I addressed this problem without having to use the coproduct. Just pick an arbitrary element of F(G), and carefully expand it out under the coproduct. Set it equal to x$\otimes$x, and see what conditions are forced on you.
Maybe that could help, a little observation on the definition of . I think it could help.
I did what Zach mentions here. I had to be careful though in drawing conclusions from setting the co-product of the arbitrary element (expanded out) equal to . The concern is that a linear combination of linearly independent pure tensors can add together to be a pure tensors also. We of course remember $$.
I found it useful to solve the equation for zero and then try to apply the “useful lemma” we saw in class.
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