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Let: $\epsilon$ be the counit and $\Delta$ the comultiplication for $A=k[t,t^{-1}]$.

If $\epsilon(t)=1$ and $\Delta(t)=t\otimes t$ does this mean that $\epsilon(t^{-1})=-1$ and $\Delta(t^{-1})=t^{-1}\otimes t^{-1}$ there appears to be no interactions in the axioms to determine if these are indeed the case. I mean, I am here mainly not sure if $\epsilon(t^{-1})=-1$ or if in general $\epsilon(t^n)=1$ for $n\in \Bbb Z$.

I guess if it is indeed an algebra homomorphism $\epsilon(t^2)=\epsilon(t)\epsilon(t)=1$ and $\epsilon(t^2t^{-1})=1\epsilon(t^{-1})\implies \epsilon(t^{-1})=1$

Why does the text only define it for $t$, does the inverse follow immediately somehow?

I imagine they actually meant to define it on the basis elements, like $\Delta(x)=x\otimes x$ for $x\in \{1,t,t^{-1},t^2,t^{-2},\cdots\}$ here, where this map is $k$-linear, and then extend by linearity.

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Remember that $\epsilon$ is a morphism, therefore

$$1 = \epsilon(1) = \epsilon(t t^{-1}) = \epsilon(t) \epsilon(t^{-1}) = \epsilon(t^{-1}) .$$

Similarly,

$$1 \otimes 1 = \Delta(1) = \Delta(t t^{-1}) = \Delta(t) \Delta(t^{-1}) = (t \otimes t) \Delta (t^{-1}) $$

whence indeed $\Delta(t^{-1}) = t^{-1} \otimes t^{-1}$.

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    Aha, thank you, funny timing with my edit! So silly of me! – 2017-02-21