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text: '[summary: Here, we prove that the [-5zv] characterises objects uniquely up to [4f4]. That is, if two objects both satisfy the universal property of the product of $A$ and $B$, then they are isomorphic objects.]\n\nRecall the [-600] of the [5zv product]:\n\n> Given objects $A$ and $B$, we define the *product* to be the following collection of three objects, if it exists: $$A \\times B \\\\ \\pi_A: A \\times B \\to A \\\\ \\pi_B : A \\times B \\to B$$ with the requirement that for every object $X$ and every pair of maps $f_A: X \\to A, f_B: X \\to B$, there is a *unique* map $f: X \\to A \\times B$ such that $\\pi_A \\circ f = f_A$ and $\\pi_B \\circ f = f_B$.\n\nWe wish to show that if the collections $(R, \\pi_A, \\pi_B)$ and $(S, \\phi_A, \\phi_B)$ satisfy the above condition, then there is an [-4f4] between $R$ and $S$. %%note: I'd write $A \\times_1 B$ and $A \\times_2 B$ instead of $R$ and $S$, except that would be really unwieldy. Just remember that $R$ and $S$ are both standing for products of $A$ and $B$.%%\n\n# Proof\n\nThe proof follows a pattern which is standard for these things.\n\nSince $R$ is a product of $A$ and $B$, we can let $X = S$ in the universal property to obtain:\n\n> For every pair of maps $f_A: S \\to A, f_B: S \\to B$ there is a unique map $f: S \\to R$ such that $\\pi_A \\circ f = f_A$ and $\\pi_B \\circ f = f_B$.\n\nNow let $f_A = \\phi_A, f_B = \\phi_B$:\n\n> There is a unique map $\\phi: S \\to R$ such that $\\pi_A \\circ \\phi = \\phi_A$ and $\\pi_B \\circ \\phi = \\phi_B$.\n\nDoing the same again but swapping $R$ for $S$ and $\\phi$ for $\\pi$ (basically starting over with the line "Since $S$ is a product of $A$ and $B$, we can let $X = R$…"), we obtain:\n\n> There is a unique map $\\pi: R \\to S$ such that $\\phi_A \\circ \\pi = \\pi_A$ and $\\phi_B \\circ \\pi = \\pi_B$.\n\nNow, $\\pi \\circ \\phi: S \\to S$ is a map which we wish to be the identity on $S$; that would get us halfway to the answer, because it would tell us that $\\pi$ is left-inverse to $\\phi$.\n\nBut we can use the universal property of $S$ once more, this time looking at maps into $S$:\n\n> For every pair of maps $f_A: S \\to A, f_B: S \\to B$ there is a unique map $f: S \\to S$ such that $\\phi_A \\circ f = f_A$ and $\\phi_B \\circ f = f_B$.\n\nLetting $f_A = \\phi_A$ and $f_B = \\phi_B$, we obtain:\n\n> There is a unique map $f: S \\to S$ such that $\\phi_A \\circ f = \\phi_A$ and $\\phi_B \\circ f = \\phi_B$.\n\nBut I claim that both the identity $1_S$ and also $\\pi \\circ \\phi$ satisfy the same property as $f$, and hence they're equal by the uniqueness of $f$.\nIndeed, \n\n- $1_S$ certainly satisfies the property, since that would just say that $\\phi_A = \\phi_A$ and $\\phi_B = \\phi_B$;\n- $\\pi \\circ \\phi$ satisfies the property, since we already found that $\\phi_A \\circ \\pi = \\pi_A$ and that $\\phi_B \\circ \\pi = \\pi_B$.\n\nTherefore $\\pi$ is left-inverse to $\\phi$.\n\nNow to complete the proof, we just need to repeat *exactly* the same steps but with $(R, \\pi_A, \\pi_B)$ and $(S, \\phi_A, \\phi_B)$ interchanged throughout.\nThe outcome is that $\\phi$ is left-inverse to $\\pi$.\n\nHence $\\pi$ and $\\phi$ are genuinely inverse to each other, so they are both isomorphisms $R \\to S$ and $S \\to R$ respectively.\n\n# The characterisation is not unique\n\nTo show that we can't do better than "characterised up to isomorphism", we show that the product is not characterised *uniquely*.\nIndeed, if $(A \\times B, \\pi_A, \\pi_B)$ is a product of $A$ and $B$, then so is $(B \\times A, \\pi'_A, \\pi'_B)$, where $\\pi'_A(b, a) = a$ and $\\pi'_B(b, a) = b$.\n(You can check that this does satisfy the universal property for a product of $A$ and $B$.)\n\nNotice, though, that $A \\times B$ and $B \\times A$ are isomorphic as guaranteed by the theorem.\nThe isomorphism is the map $A \\times B \\to B \\times A$ given by $(a,b) \\mapsto (b,a)$.',
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