Differential of a morphism of abelian varieties I am reading the lecture notes of J.S. Milne on Abelian varieties and I got stuck at some point. 
Let $\alpha,\beta\colon X\rightarrow Y$ be homomorphisms of abelian varieties $X$ and $Y$. Then for the differential maps we have the equality: $d\alpha_0+d\beta_0=d(\alpha+\beta)_0$ as maps between tangent spaces $T_0X$ and $T_0 Y$ (By abuse of notation, $0$ is the identity element of both $X$ and $Y$).
Over $\mathbb{C}$, it is clear by GAGA, but how would one show this equality over an arbitrary field?
 A: We may as well replace $X$ and $Y$ with their base change to $\bar{k}$, say $\overline{X}$ and $\overline{Y}$. But, then, we may as well replace them with their classic varieties $\overline{X}\left(\bar{k}\right)$ and $\overline{Y}\left(\bar{k}\right)$. Thus, I assume that $X$ and $Y$ are classic varieties over the algebraically closed field $k$.
It's common knowledge that for $(0,0)\in Y\times Y$ we have a decomposition of $T_{(0,0)}(Y\times Y)$ as $T_0 Y\oplus T_0 Y$ where we identify the copies of $T_0 Y\subseteq T_{(0,0)}(Y\times Y)$ by the derivative of the inclusions $\iota_j:Y\hookrightarrow Y\times Y$ sending $x\mapsto (x,0)$ and $x\mapsto (0,x)$
Now, with this, we claim that $d\mu:T_{(0,0)}(Y\times Y)\to T_0 Y$, where $\mu:Y\times Y\to Y$ is the addition map, is nothing but $(v,w)\mapsto v+w$, when we identify $T_{(0,0)}(Y\times Y)$ with $T_0 Y\oplus T_0 Y$. To see this, we merely note that
$$d\mu(v,w)=d\mu(d\iota_1(v)+d\iota_2(w))=d\mu(d\iota_1(v))+d\mu(d\iota_2(w))$$
But, 
$$d\mu\circ d\iota_j=d(\mu\circ \iota_j)=d\text{id}=\text{id}$$
and so
$$d\mu(v,w)=d\mu(d\iota_1(v))+d\mu(d\iota_2(w))=v+w$$
as desired.
Now, consider two maps $\alpha,\beta:X\to Y$. Then, $\alpha+\beta$ is, by definition, the composition
$$X\xrightarrow{(\alpha,\beta)}Y\times Y\xrightarrow{\mu}Y$$
Thus, we have
$$d(\alpha+\beta)(v)=d\mu(d(\alpha,\beta)(v))=d\mu(d\alpha(v),d\beta(v))=d\alpha(v)+d\beta(v)$$
which is what we wanted.
