In section II.6 "the theorem of the cube I" of Mumford's "Abelian Variety" book, Mumford introduced the notion of functor of order $n$. Here is part of the remark immediately below the statement of theorem.
Let $T$ be a contravariant functor on the category of complete varietties into the Category $\underline{\mathrm{Ab}}$ of abelian groups. Let $X_0,\ldots,X_n$ be any system of complete varieties, $x_i^0$ a bse point of $X_i$ and let $\pi_i: X_0\times\ldots\times X_n\to X_0\times\ldots\times \widehat{X_i}\times \ldots X_n$ ($\widehat{X_i}$ indicating the omission of the $i$-th factor $X_i$) be the projection map, and $\sigma_i: X_0\times\ldots\times \widehat{X_i}\times \ldots X_n \to X_0\times\ldots \times X_n$ the inclusion map using the base point.
Consider the homomorphisms
$$\alpha_T^n: \prod_{i=0}^n T(X_0\times\ldots\times \widehat{X_i}\times \ldots X_n) \to T(X_0\times\ldots \times X_n),$$ $$ \beta_T^n: T(X_0\times\ldots \times X_n)\to \prod_{i=0}^n T(X_0\times\ldots\times \widehat{X_i}\times \ldots X_n),$$ defined in the natural way.
One then prove by an easy induction on $n$ that we have natrual splitting $$T(X_0\times\ldots \times X_n)=\mathrm{Im}\alpha\oplus \ker \beta.$$ The functor is said to be order $n$ if $\alpha $ is surjective or equivalently, $\beta$ is injective.
Now if $T_i, i=1,2,3$ are contravariant functors on complete varieties into $\underline{\mathrm{Ab}}$, and $T_1\to T_2$ and $T_2\to T_3$ are natrual transformations such that $T_1\to T_2\to T_3$ is an exact sequence, and if $T_1$ and $T_3$ are of order $n$ so is $T_2$, as follows from the exactness of
$$0=\ker\beta^n_{T_1}(X_0\times\ldots \times X_n)\to \ker\beta^n_{T_2}(X_0\times\ldots \times X_n) \to\ker\beta^n_{T_3}(X_0\times\ldots \times X_n)=0.$$
There are two things that I am having troube with. First, I have tried a couple times to work out this "easy" induction but have not much success so far. The base case is easy. For $n=2$, I got $\mathrm{Im}{\alpha}\cap \ker\beta=0$ part. But not much else.
Secondly, I don't see where the exactness of this sequence involving the $\ker\beta_{T_i}$'s come from. It feels to me that one is claiming that if we have a commutative diagram with exact rows, then the kernel of the vertical maps fit into a sequence that's exact in the middle, which is certainly not true, as one can easily take $B'=C'=0$ and produce counter examples.
$$\begin{array}{cccc} A&\to & B & \to & C\\ \downarrow& & \downarrow & & \downarrow\\ A' &\to & B' & \to & C'\\ \end{array}$$