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  "body": "\n## Statement\n\nLet $X$ and $Y$ be topological spaces with singular chain complexes $C_\\bullet(X)$ and $C_\\bullet(Y)$.\nLet $C_\\bullet(X) \\otimes C_\\bullet(Y)$ be the tensor product chain complex, with $(C_\\bullet(X) \\otimes C_\\bullet(Y))_n = \\bigoplus_{i + j = n} C_i(X) \\otimes_{\\Z} C_j(Y)$ and $\\partial(a \\otimes b) = \\partial a \\otimes b + (-1)^i a \\otimes \\partial b$ for $a \\in C_i(X)$.\nThen there are chain maps\n$$C_\\bullet(X \\times Y) \\rightleftarrows C_\\bullet(X) \\otimes C_\\bullet(Y),$$\nnatural in $X$ and $Y$, that are inverse to each other up to chain homotopy.\nIn particular, $H_n(X \\times Y) \\cong H_n\\big(C_\\bullet(X) \\otimes C_\\bullet(Y)\\big)$ for every $n$.\n\n## Remarks\n\nA standard choice of maps is the Alexander–Whitney map from $C_\\bullet(X \\times Y)$ and the shuffle map into it.\nThe existence and uniqueness up to homotopy of such natural maps follows from the method of acyclic models.\nBoth functors are free on the models $(\\Delta^p, \\Delta^q)$ and acyclic on them, because $\\Delta^p \\times \\Delta^q$ is contractible.\n\nExample: with the algebraic Künneth theorem, the theorem gives the topological Künneth formula.\nOver a field $F$, $H_n(X \\times Y; F) \\cong \\bigoplus_{i + j = n} H_i(X; F) \\otimes_F H_j(Y; F)$.\nFor the torus $S^1 \\times S^1$ this gives $H_1 \\cong F^2$ and $H_2 \\cong F$.\n\n## Explanations\n\n### L1 human\n\nThe product of two spaces, such as a square from two intervals, can be cut into pieces in two ways.\nOne can use simplices of the product, or products of a simplex of each factor.\nThe theorem says that both ways give the same homology.\nSo the homology of a product can be computed from the chains of the two factors.\n\n### L2 human\n\nThe theorem is the bridge between the homology of a product and the algebra of tensor products of chain complexes.\nCombined with the Künneth theorem of homological algebra, it computes the homology of products.\nThe Alexander–Whitney map, composed with the diagonal $X \\to X \\times X$, also gives the cup product on singular cohomology.\n",
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