Notation: Let $\pi$'s denote canonical projections, $j$'s denote canonical injections, given a pair of maps $X \overset{f_1}{\to} Y_1$ and $X \overset{f_2}{\to} Y_2$ denote the unique map $X \overset{f}{\to} Y_1 \times Y_2$ s.t. $\pi_1 \circ f = f_1$ and $\pi_2 \circ f = f_2$ by $\langle f_1, f_2 \rangle$, then given $X_1 \overset{f_1}{\to} Y_1$ and $X_2 \overset{f_2}{\to}Y_2$ let $X_1 \times X_2 \overset{f_1 \times f_2}{\to} Y_1 \times Y_2$ denote $\langle f_1 \circ \pi_1, f_2 \circ \pi_2 \rangle$.
Question: In a category where sums and products exist, when, if ever, does the object $A \times (B_1 + B_2)$ satisfy the following universal property?
- For every pair of maps $A \overset{f_1}{\to} C_1$, $B_1 + B_2 \overset{f_2}{\to} C_2$, there exists exactly one map $A \times (B_1 + B_2) \overset{f}{\to} C_1 \times C_2$ such that $f \circ (1_A \times j_1) = f_1 \times (f_2 \circ j_1)$ and $f \circ (1_A \times j_2)= f_1 \times (f_2 \circ j_2)$.
In other words, given any $A \times (B_1 + B_2) \overset{g}{\to} C_1 \times C_2$ with $g \circ (1_A \times j_1) = f_1 \times (f_2 \circ j_1)$ and $g \circ (1_A \times j_2)= f_1 \times (f_2 \circ j_2)$, it follows that $g = f_1 \times f_2 = f$.
A seemingly necessary condition for this is that the canonical projections $\pi_i$ be epimorphisms, and a sufficient condition for that necessary condition is that the category has a zero object according to this related question. The above also seems like it would be a special case of something like:
- For every pair of maps $A_1 \overset{f_1}{\to} C_1$, $A_2 \overset{f_2}{\to} C_2$, there is exactly one map $A_1 \times A_2 \overset{f}{\to} C_1 \times C_2$ such that $f=f_1 \times f_2$, and $f = g_1 \times g_2$ implies that $f_1 = g_1$ and $f_2 = g_2$.
(Since we have in the case that $A_2 = B_1 + B_2$ by the universal product of the sum that $g \circ j_1 = f_2 \circ j_1$ and $g \circ j_2 = f_2 \circ j_2$ implies that $g = f_2$.) Obviously one such map exists, so what would need to be shown is that $g_1 \times g_2 = \langle g_1 \circ \pi^A_1, g_2 \circ \pi^A_2 \rangle = \langle f_1 \circ \pi^A_1, f_2 \circ \pi^A_2 \rangle = f_1 \times f_2$ implies that $f_1 = g_1$ and $f_2 = g_2$. By the universal property of the product I think this already implies that $f_1 \circ \pi_1^A = g_1 \circ \pi_1^A$ and $f_2 \circ \pi_A^2 = g_2 \circ \pi_A^2$, so if $\pi_1^A$ and $\pi_2^A$ are epic then we're done.
Context: I am trying to show that in a Cartesian closed category with sums and products, the standard map $A \times B_1 + A \times B_2 \overset{s}{\to} A \times (B_1 + B_2)$ has an inverse, call it $t$, defined using the universal property of the exponential object.
I have been able to show that $t \circ s = 1_{A \times B_1 + A \times B_2}$ quite easily, by showing that $(t \circ s) \circ J_1 = J_1$ and $(t \circ s) \circ J_2 = J_2$ (where $J_1$ and $J_2$ are the corresponding injections $\to A \times B_1 + A \times B_2$).
However, I have not been able to show that $s \circ t = 1_{A \times (B_1 + B_2)}$. I had tried to do this by showing that $\pi_1 \circ (s \circ t) = \pi_1$ and $\pi_2 \circ (s \circ t) = \pi_2$, but the left expressions don't seem to simplify at all. However, it is quite easy to show that $(s \circ t) \circ (1_A \times j_1) = 1_A \times j_1$ and $(s \circ t) \circ (1_A \times j_2) = 1_A \times j_2$. ($j_i$ denote the injections $B_1 + B_2$.) So ideally if it is true that $1_{A \times (B_1 + B_2)}$ is the unique $f$ such that $f \circ (1_A \times j_1) = 1_A \times j_1$ and $f \circ (1_A \times j_2) = 1_A \times j_2$, and if I can show that this is true, then I would be done. It seems intuitively plausible, but I'm stuck trying to come up with either a counterexample or a proof (hence the above question).