Functor of order $n$ as in Mumford's Abelian varieties 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.   
$$\begin{array}{cccc}
A&\to & B & \to &  C\\
\downarrow& & \downarrow & & \downarrow\\
A' &\to & B' & \to &  C'\\
\end{array}$$
 A: I got stuck on this claim too. The following proof doesn't use induction, so it may not be exactly what Mumford had in mind, but it works. The birds-eye-view idea is that we modify his $\alpha$ and $\beta$ in a way that doesn't change the image (resp. kernel) but deals with the fact that cleaning up junk in the $n-1$ position creates junk in the $n-2$ position.
The result doesn't end up using much about complete varieties, so we will abstract away a bit.
We have $n+1$ objects $X_0, \ldots, X_n$ in some category with products.
Write $[n] = \{0, 1, \ldots, n\}$. If $S$ is a subset of $[n]$, write $X_S$ for $\prod_{i \in S} X_i$.
For $S \supset T$, we have the projection maps
$$
\pi_{S, T}: X_S \to X_T
$$
and we suppose that there are also maps
$$
\sigma_{T, S}: X_T \to X_S
$$
satisfying the following axioms:


*

*$\sigma_{T_2, S} \circ \sigma_{T_1, T_2} = \sigma_{T_1, S}$ when $T_1 \subset T_2 \subset S$.

*$\sigma_{S, S} = \text{id}$.

*$\pi_{S, T} \circ \sigma_{T, S} = \text{id}$.

*$\sigma_{T, [n]} \circ \pi_{[n], T} \circ \sigma_{S, [n]} = \sigma_{S \cap T, [n]} \circ \pi_{S, S \cap T}$

*$\pi_{[n], T} \circ \sigma_{S,[n]} \circ \pi_{[n], S} = \sigma_{S \cap T, T} \circ  \pi_{[n], S\cap T}$.


These axioms are satisfied in Mumford's situation where the $\sigma$ are inclusion of faces into the product along fixed basepoints.
(His $\pi_i$ is our $\pi_{[n], [n] \setminus \{i\}}$ and his $\sigma_i$ is our $\sigma_{[n] \setminus \{i\}, [n]}$).
We will drop $[n]$ from the subscripts, so $\pi_T$ means $\pi_{[n], T}$ and $\sigma_T$ means $\sigma_{T, [n]}$.
If $F$ is a functor valued in abelian groups, Mumford's map
$$
\alpha: \prod_{\#T = n-1} F(X_T) \to F(X_{[n]})
$$
is given by $(x_T) \mapsto \sum_T \pi_T^*(x)$, and his map
$$
\beta:  F(X_{[n]}) \to \prod_{\#T = n-1} F(X_T)
$$
is given by $x \mapsto (\sigma_T^*(x))_T$.
We are going to modify Mumford's maps by dropping the restriction that $\#T = n-1$ and introducing some signs. To this end, set
$$
\alpha_{\max}: \prod_T F(X_T) \to F(X_{[n]}), \quad (x_T) \mapsto (-1)^{n + 1} \sum_{T \subsetneq [n]} (-1)^{\#T} \pi_T^*(x_T).
$$
and
$$
\beta_{\max}: F(X_{[n]}) \to \prod_{T \subsetneq [n]} F(X_T), \quad x \mapsto (\sigma_T^* x).
$$
It follows from the axioms that the images of $\alpha$ and $\alpha_{\max}$ are the same, as are the kernels of $\beta$ and $\beta_{\max}$.
So we may as well forget Mumford's maps and just write $\alpha$, $\beta$ for $\alpha_{\max}$ and $\beta_{\max}$.
We will check that (for our maps $\alpha$, $\beta$) one has $\alpha\beta\alpha = \alpha$ and $\beta\alpha\beta = \beta$.
Spotting this, the result follows formally:
There is a tautological exact sequence
$$
0 \to \text{ker}(\alpha\beta) \to F(X_{[n]}) \to \text{Im}(\alpha\beta) \to 0
$$
but $\text{ker}(\alpha\beta) = \text{ker}(\beta)$ since $\alpha\beta x = 0 \Rightarrow \beta x = \beta\alpha\beta x = 0$,
and $\text{Im}(\alpha\beta) = \text{Im}(\alpha)$ since $\text{Im}(\alpha) \supset \text{Im}(\alpha\beta) \supset \text{Im}(\alpha\beta\alpha) = \text{Im}(\alpha)$
The exact sequence splits since any
$$
x = x - \alpha\beta x + \alpha\beta x
$$
where the sum of the first two summands is in $\text{ker}(\beta)$ and the last is in $\text{im}(\alpha)$,
and this formula shows the splitting is natural in $F$.
It remains to compute the claimed equalities.
For $\beta\alpha\beta(x)$, which is a tuple indexed by proper subsets of $[n]$, the term in the spot indexed by $S$ is
$$
(-1)^{n+1}\sum_{T \neq [n]} (-1)^{\#T} \sigma^*_S\pi^*_T\sigma^*_T(x).
$$
Using the axioms, the summand corresponding to $T$ only depends on $T \cap S$ and is given by
$$
 \pi_{S, S\cap T}^*\sigma_{S \cap T}^*(x)
$$
Given $S' \subset S$, let $m$ denote $n - \#T$.
The $T \subset [n]$ such that $T \cap S = S'$ are given by unioning $S'$ with subsets of $[n] \setminus S$.
For $i = 0, \ldots, m$, there are ${m \choose i}$ such $T$ of cardinality $\# S'$ + $i$, except in the case
where $S' =S$ and $i = m$, where there are $0$ such subsets instead of $1$ (because $[n]$ itself does not
contribute to the sum).
The summands with $T \cap S = S' \neq S$ therefore contribute
$$
(-1)^{n+1}\sum_{i=0}^m (-1)^{\#S + i} {m \choose i} \sigma_{S \cap T}^* \pi_{S, S'}^*x = 0.
$$
The summands with $T \cap S = S$ contribute
$$
(-1)^{n+1} \sum_{i=0}^{m-1} (-1)^{\#S + i} {m \choose i} \sigma_S^* x = -(-1)^{n+1} (-1)^n \sigma_S^* x = \sigma_S^* x,
$$
which is the term of $\beta(x)$ in the spot indexed by $S$, finishing the computation.
The $\alpha \beta \alpha = \alpha$ proof is similar and perhaps they can be combined
(although note that $\alpha \beta \neq \text{id}$ and $\beta \alpha \neq \text{id}$).
Given a tuple $(x_T)$, the component in spot $S$ of $\beta \alpha (x_T)$ is
$$
(-1)^{n+1} \beta(\sum_T (-1)^{\#T} \sigma_S^*\pi_T^* x_T)
$$
and so $\alpha \beta \alpha (x_T)$ is
$$
\sum_S \sum_T (-1)^{\#S + \#T} \pi^*_S\sigma^*_S\pi^*_T x_T.
$$
The summand of the coefficient of $x_T$ corresponding to $S$ only depends on $T \cap S$ and is given by
$$
\pi_{S \cap T}^* \sigma_{S \cap T, T}^* (x_T).
$$
Adding over $S$ with $S \cap T = T' \subsetneq T$, the contribution to the coefficient of $x_T$ is zero as in the other proof.
For the $S$ with $S \cap T = T$, we are missing $S = [n]$ so instead get
$$
-(-1)^{n+\#T} \pi_T^* x_T.
$$
The double sum thus simplifies to
$$
(-1)^{n+1} \sum_T (-1)^{\#T}\pi_T^* x_T = \alpha (x_T)
$$
as desired.
===
Your second question follows from the naturality: given an exact sequence of functors, the morphisms must respect the decomposition into $\text{ker}(\beta) \oplus \text{im}(\alpha)$. This follows from our explicit formula for the decomposition. So the $\text{ker}(\beta)$ and $\text{im}(\alpha)$ pieces of the exact sequence are separately exact.
===
The signs in the proof may seem to come out of nowhere but are natural if you consider the picture geometrically. You have some object on the cube and you want to know whether it came from the $n-1$ faces, so you subtract off the bits that came from the $n-1$ faces. But oh no, that's done double damage on the $n-2$ faces, so you'd better add them back, etc.
