Linear categories in Lawvere's Conceptual Mathematics In Lawvere's Conceptual Mathematics, linear categories (apparently called additive categories elsewhere) are defined as seen in this paragraph:
Linear category definition
Lawvere proceeds to define a 'product' of two matrices in a linear category as follows:
Matrix product definition
And then makes the following claim:
Map identity with f,g and h
I've tried in vain to prove Lawvere's claim (following his definitions) that the product of these two matrices must indeed have the form that he stated, with the first row being $1,h$ and the second row being $0,1$. No additional explanation is provided in the text.
Why must this be true? 
 A: The product $X\times Y$ implicitly comes with two fixed projections $\pi_X,\pi_Y$, and these play an implicit role in the definitions. 
Dually, the coproduct $X+Y$ comes equipped with inclusions $\iota_X,\iota_Y$.
First, observe that (writing composition to the right), 
$$\underset{X+Y\,\to\, X\times Y\,\to\, X}{\pmatrix{a&b\\c&d}\pi_X}\ =\ \pmatrix{a\\c}$$
and similarly, composing it by $\pi_Y$ yields the second column. 
Consequently, as $1_{X\times Y}=\alpha\,\pmatrix{1&0\\0&1}$, we have 
$$ \pi_X\ =\ \alpha\,\pmatrix{1&0\\0&1}\pi_X\ =\underset{X\times Y\to X+Y\to X}{\qquad \alpha\,\pmatrix{1\\0}}\,. $$
Dual statements hold for rows and the $\iota$'s.
From here, we can evaluate the 3 required elements like:
$$\iota_Y\cdot\,\pmatrix{1&f\\0&1}\alpha\pmatrix{1&g\\0&1}\,\cdot\pi_X\ =\ 
\pmatrix{0&1}\,\alpha\,\pmatrix{1\\0}\ =\ \pmatrix{0&1}\pi_X\ =\ 0\,.$$
A: This answer seems to be exactly the same as Berci's except worse. But I'll write the idea using Lawvere's notation so that maybe it is more clear to someone using the book (like me) learning this for the first time.
If the map $$ A \times B \overset{\alpha_{AB}}{\to} A +B $$ denotes the inverse of the "identity matrix"
$$\begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} $$
then we have the following identities:

$$ \alpha_{AB} \circ \langle 1_A, 0_{AB} \rangle = j_{A+B}^1 \,, \quad \alpha_{AB} \circ \langle  0_{BA}, 1_B \rangle = j_{A+B}^2 \,, \quad \begin{cases} 1_A \\ 0_{BA} \end{cases} \circ \alpha_{AB} = \pi_{A \times B}^1 \,, \quad \begin{cases} 0_{AB} \\ 1_B \end{cases} \circ \alpha_{AB} = \pi_{A \times B}^2 \,.$$

The proofs follow from the definition of $\alpha_{AB}$, namely:
$$ \alpha_{AB} \circ \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases}  = 1_{A+B} \implies j_{A+B}^1 = \alpha_{AB} \circ \left( \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^1 \right) = \alpha_{AB} \circ \langle 1_A, 0_{AB} \rangle  $$
$$ \alpha_{AB} \circ \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases}  = 1_{A+B} \implies j_{A+B}^2 = \alpha_{AB} \circ \left( \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^2 \right) = \alpha_{AB} \circ \langle  0_{BA} , 1_B \rangle  $$
$$ \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} \circ \alpha_{AB} = 1_{A \times B} \implies \pi_{A \times B}^1 = \left( \pi_{A \times B}^1  \circ \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} = \left(  \begin{cases} \pi_{A \times B}^1 \circ \langle 1_A, 0_{AB}\rangle \\  \pi_{A \times B}^1 \circ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} = \begin{cases} 1_A \\ 0_{BA} \end{cases} \circ \alpha_{AB} $$
$$ \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} \circ \alpha_{AB} = 1_{A \times B} \implies \pi_{A \times B}^2 = \left( \pi_{A \times B}^2  \circ \begin{cases} \langle 1_A, 0_{AB}\rangle \\  \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} = \left(  \begin{cases} \pi_{A \times B}^2 \circ \langle 1_A, 0_{AB}\rangle \\  \pi_{A \times B}^2 \circ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} = \begin{cases} 0_{AB} \\ 1_B \end{cases} \circ \alpha_{AB} $$ $\triangle$
With that under our belt, the result is fairly straightforward:
$$ \begin{bmatrix} 1_A & f \\ 0_{BA} & 1_B \end{bmatrix} \cdot \begin{bmatrix} 1_A & g \\ 0_{BA} & 1_B \end{bmatrix} = \begin{cases} \langle 1_A , g \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \circ \alpha_{AB} \circ \begin{cases} \langle 1_A , f \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} =: \begin{bmatrix} h_{AA} & h_{AB} \\ h_{BA} & h_{BB} \end{bmatrix}  \,.$$
Therefore we have that:
$$ h_{AA} = \left( \pi_{A+B}^1 \circ \begin{cases} \langle 1_A , g \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} \circ  \left(  \begin{cases} \langle 1_A , f \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^1 \right) = \begin{cases} 1_A \\ 0_{BA} \end{cases} \circ \alpha_{AB} \circ \langle 1_A, f \rangle = \pi_{A\times B}^1 \circ \langle 1_A, f \rangle = 1_A $$
$$ h_{BA} = \left( \pi_{A+B}^1 \circ \begin{cases} \langle 1_A , g \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} \circ  \left(  \begin{cases} \langle 1_A , f \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^2 \right) = \begin{cases} 1_A \\ 0_{BA} \end{cases} \circ \alpha_{AB} \circ \langle 0_{BA}, 1_B \rangle =\begin{array}{c} \pi_{A\times B}^1 \circ \langle 0_{BA}, 1_B \rangle \\ \begin{cases}1_A \\ 0_{BA} \end{cases} \circ j_{A+B}^2 \end{array} = 0_{BA} $$
$$ h_{BB} = \left( \pi_{A+B}^2 \circ \begin{cases} \langle 1_A , g \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} \circ  \left(  \begin{cases} \langle 1_A , f \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^2 \right) = \begin{cases} g \\ 1_{B} \end{cases} \circ \alpha_{AB} \circ \langle 0_{BA}, 1_B \rangle = \begin{cases} g \\ 1_{B} \end{cases}  \circ j_{A+B}^2 = 1_B$$
We even get as a bonus an explicit form for $f+g$ (although to be honest I have not yet found this helpful for solving the subsequent Exercise 1, which is why I wound up on Math.SE in the first place):
$$f+g := h_{AB} = \left( \pi_{A+B}^2 \circ \begin{cases} \langle 1_A , g \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} \circ  \left(  \begin{cases} \langle 1_A , f \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^1 \right) = \begin{cases} g \\ 1_B \end{cases} \circ \alpha_{AB} \circ \langle 1_A, f \rangle \,. $$
