Proof that the given sequence is exact Let $K\rightarrow M\rightarrow N\rightarrow 0$ be a short exact sequence of modules over a commutative ring $R$, then prove that 
$T(N)\otimes K\otimes T(N)\rightarrow T(N)\rightarrow T(M)\rightarrow 0$ is a short exact sequence, where $T(-)$ is the tensor algebra of the given module.
I'm reading through Eisenbud's Commutative Algebra with a View Twoards Algebraic Geometry, and in in one of the appendices, it seems to assume this sequence is exact to prove that another sequence is exact. I understand how the sequence is exact at $T(M)$ since $\otimes$ is an exact functor, but I'm not understanding how to define the maps induced but $T(-)$. If I understood the maps, I could possibly deduce the exactness at $T(N)$ on my own. 
 A: (The given exact sequence is not short, at least for me, it can maybe be called a "left incomplete short exact sequence.) The given exact sequence is not the one in loc. cit, page 571, Proposition A2.2, (d). (There is no map $TN\to TM$ in the given setting.) So i have to restate:
Given: $K\to N\to M\to 0$ is an exact sequence of $R$-modules. We show that
$$ TN\otimes K\otimes TN\to TN\to TM\to 0$$
is exact.
For this, let us start with two exact sequences in the category of $R$-modules, 
$$
\begin{aligned} 
&K_1\to N_1\to M_1\to 0\ ,\\
&K_2\to N_2\to M_2\to 0\ ,
\end{aligned} 
$$
and show 
$$
\color{blue}{N_1\otimes K_2\ \oplus\  K_1\otimes N_2}
\qquad\to\qquad \color{red}{N_1\otimes N_2}
\qquad\to\qquad \color{red}{M_1\otimes M_2}
\qquad\to\qquad 0
$$
is exact.
Consider for the last the diagram with exact rows and columns ($\otimes$ being right exact):
$\require{AMScd}$
\begin{CD}
    0  @<<< M_1\otimes K_2@<<< \color{blue}{N_1\otimes K_2}@<<< K_1\otimes K_2\\
    @. @VVV @VVV @VVV \\
    0  @<<< M_1\otimes N_2@<<< \color{red}{N_1\otimes N_2}@<<< \color{blue}{K_1\otimes N_2}\\
    @. @VVV (*) @VVbV @VVV \\
    0  @<<< \color{red}{M_1\otimes M_2}@<<a< N_1\otimes M_2@<<< K_1\otimes M_2\\
    @. @VVV @VVV @VVV \\
       @. 0 @. 0 @. 0 
\end{CD}
We are searching for the kernel of the surjective diagonal map $M_1\otimes M_2\leftarrow N_1\otimes N_2$, $ab$, between the red entries. We will show it is the sum of the blue entries. This follows from the exactness in
$$ 0
\leftarrow \operatorname{Coker} a
\leftarrow \operatorname{Coker} ab
\leftarrow \operatorname{Coker} b
$$
or from the corresponding diagram chase, here explicitly as follows. 
Start with 
$\color{red}{\sum n_1\otimes n_2}$
(in slightly abusive notation)
mapped to zero diagonally via $ab$. 
We want to show that this sum is of the shape 
$\color{red}{\sum n_1\otimes n_2}\sim
\color{blue}{\sum n'_1\otimes k_2}
\color{red}{+}
\color{blue}{\sum k_1\otimes n'_2}
$
where we use "lower case notations" for elements in capital case objects.
The $\sim$ means that the blue entries have to be mapped first in the red world, where the plus finally makes sense.
Then $b$ maps this starting element in an element, denoted in the same spirit $\sum n_1\otimes m_2$. It is mapped to zero horizontally via $a$, so it comes from an element $\sum k_1\otimes m'_2$. We lift this in the blue entry above, get an element
$\sum k_1\otimes n'_2$.
And map this element back in the middle, getting an element $\sum n'_1\otimes n'_2$, say. Now consider the difference 
$\sum n_1\otimes n_2-\sum n'_1\otimes n'_2$.
It is vertically mapped via $b$ to zero. So it comes from the second blue entry above, say from $\sum n''_1\otimes k_2$. Then
$$
\begin{aligned}
&\color{blue}{\sum n''_1\otimes k_2}
\oplus
\color{blue}{\sum k_1\otimes n'_2}
\\
&\qquad\qquad\qquad
\to
\left(
\sum n_1\otimes n_2-\sum n'_1\otimes n'_2
\right)
+
\left(
\sum n'_1\otimes n'_2
\right)
\\
&\qquad\qquad\qquad
=\sum n_1\otimes n_2\ .
\end{aligned}
$$
This show the special form for the element "in the kernel", as an element "in the image", i.e. the needed exactity.
The above generalizes inductively. Given:
$$
\begin{aligned} 
&K_1\to N_1\to M_1\to 0\ ,\\
&K_2\to N_2\to M_2\to 0\ ,\\
&\vdots\qquad\vdots\qquad\vdots\qquad\vdots\qquad\vdots\qquad\\ 
&K_r\to N_r\to M_r\to 0\ ,\\
\end{aligned} 
$$
we have the exact sequence:
$$
\color{blue}{\bigoplus \dots N_{j-1}\otimes K_j\otimes N_{j+1}\dots}
\to \color{red}{N_1\otimes N_2\otimes\dots\otimes N_r}
\to \color{red}{M_1\otimes M_2\otimes\dots\otimes M_r}
\to 0\ .
$$
We need the above for equal values of the $K$'s, of the $N$'s, and of the $M$'s, explicitly
$$
\color{blue}{\bigoplus_{p+q+1=r} T^p N\otimes K\otimes T^qN}
\to \color{red}{T^r N}
\to \color{red}{T^r M}
\to 0\ .
$$
