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?

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    $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ – José Carlos Santos Sep 9 '18 at 11:12

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: $$\iota_Y\cdot\,\pmatrix{1&f\\0&1}\alpha\pmatrix{1&g\\0&1}\,\cdot\pi_X\ =\ \pmatrix{0&1}\,\alpha\,\pmatrix{1\\0}\ =\ \pmatrix{0&1}\pi_X\ =\ 0\,.$$

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    $\begingroup$ I can follow your reasoning, but I am curious if you can generalize this to show that $\pmatrix{a&b}\alpha \pmatrix{x\\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above). $\endgroup$ – user97678 Sep 10 '18 at 7:36
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    $\begingroup$ Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$\pmatrix{a&b}\pmatrix{c\\d} =\pmatrix{a&0}\pmatrix{c\\d} +\pmatrix{0 & b}\pmatrix{c\\d} =ac+bd$$ $\endgroup$ – Berci Sep 10 '18 at 7:46
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    $\begingroup$ I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things. $\endgroup$ – user97678 Sep 10 '18 at 8:20
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    $\begingroup$ I can compute $\pmatrix{1&(0,b)\\0&1} \pi_Y = \pmatrix{b \\ \pi_Y}$ but I don't know how to cancel out the $\alpha$ in the multiplication. $\endgroup$ – user97678 Sep 10 '18 at 17:52
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    $\begingroup$ Hmm.. Maybe the easiest way would be, if we could already fix all the $\pi_X,\pi_Y$ and $\iota_X,\iota_Y$ so that $\pmatrix{1_X&0\\0&1_Y}$ - hence $\alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $X\times Y\cong X+Y$, the object $X\times Y$ itself can serve as the coproduct. (Basically we put the $\alpha$'s in the inclusions $\iota_X,\iota_Y$.) $\endgroup$ – Berci Sep 10 '18 at 21:38

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):

$$f+g := h_{AB} = \left( \pi_{A+B}^2 \circ \begin{cases} \langle 1_A , g \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \right) \circ \alpha_{AB} \circ \left( \begin{cases} \langle 1_A , f \rangle \\ \langle 0_{BA}, 1_B \rangle \end{cases} \circ j_{A+B}^1 \right) = \begin{cases} g \\ 1_B \end{cases} \circ \alpha_{AB} \circ \langle 1_A, f \rangle \,. $$


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