Let $p:\mathbb{G}\_m\to \operatorname{Spec} k$ be the structure map, and let $T$ be an algebraic $k$-torus viewed as an étale sheaf over $k$. Why is the cokernel of the canonical map $T\to p\_\*p^\*T$ canonically isomorphic to the cocharacter lattice $L$ of $T$?


If $\operatorname{Spec}A$ is affine scheme, it seems to me that $p^\*T=T\times \mathbb{G}\_m$, so $$p\_\*p^\*T(A)=p\_\*((T\times \mathbb{G}\_m))(A)=T(A[t^{\pm 1}])\times \mathbb{G}\_m(A[t^{\pm 1}]).$$ Say that $A=K$ is a field. Then $$p\_\*p^\*T(K)=L\oplus\mathbb{Z},$$ but why is the image of $T(K)$ equal to $\mathbb{Z}$?



Cokernel of map of étale sheaves
There is a small issue with your computation related to $\mathbb G\_m$-linearity. By the $\operatorname{Gal}(k^{\text{sep}}/k)$-module view on étale sheaves on $\operatorname{Spec} k$, it suffices to understand the $k^{\text{sep}}$-points. We compute
\begin{align\*}
p\_\*p^\*T(k^{\text{sep}}) &= p^\*T(\mathbb G\_{m,k^{\text{sep}}}) = \operatorname{Hom}\_{\mathbb G\_m}(\mathbb G\_{m,k^{\text{sep}}}, T \times \mathbb G\_m) \\
&= \operatorname{Hom}\_k(\mathbb G\_{m,k^{\text{sep}}}, T) = \operatorname{Hom}\_{k^{\text{sep}}}(\mathbb G\_{m,k^{\text{sep}}}, T\_{k^{\text{sep}}})\\
&=\operatorname{Hom}\_{k^{\text{sep}}}(k^{\text{sep}}[T], k^{\text{sep}}[x,x^{-1}]).
\end{align\*}
The image of the unit $T \to p\_\*p^\*T$ will be the subgroup $\operatorname{Hom}\_{k^{\text{sep}}}(k^{\text{sep}}[T],k^{\text{sep}})$. If¹ $T\_{k^{\text{sep}}} \cong \mathbb G\_m^n$, then we get
\begin{align\*}
\operatorname{Hom}\_k(k[T],k^{\text{sep}}[x,x^{-1}]) &= \operatorname{Hom}\_{k^{\text{sep}}}(k^{\text{sep}}[t\_1^{\pm 1}, \ldots, t\_n^{\pm 1}],k^{\text{sep}}[x,x^{-1}])\\
&= \Big((k^{\text{sep}})^\times \times \mathbb Z\Big)^n,\\
\operatorname{Hom}\_k(k[T],k^{\text{sep}}) &= \operatorname{Hom}\_{k^{\text{sep}}}(k^{\text{sep}}[t\_1^{\pm 1}, \ldots, t\_n^{\pm 1}],k^{\text{sep}}) \\
&= \Big((k^{\text{sep}})^\times\Big)^n,
\end{align\*}
so the quotient is just $\mathbb Z^n$. The Galois structure is that of the cocharacter lattice of $T$. $\square$


(In the above, all $\operatorname{Hom}$ sets indicate morphisms of rings or schemes, not of group schemes.)




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¹Every torus splits over $k^{\text{sep}}$. See for example Lemma B.1.5 of [Conrad's notes](http://math.stanford.edu/~conrad/papers/luminysga3.pdf).


