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Suppose $\mathcal{A}$ and $\mathcal{B}$ are categories with products, and $T$ a functor between them. If $X$ and $Y$ are objects in $\mathcal{A}$, what does it mean when we say there is a natural morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$?

In $\mathcal{A}$, we have the product $X\times Y$, with corresponding morphisms $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. Under $T$, we get a diagram of objects in $\mathcal{B}$ of morphisms $T(\pi_1):T(X\times Y)\to T(X)$ and $T(\pi_2):T(X\times Y)\to T(Y)$.

Since products exist in $\mathcal{B}$, we have a product $(T(X)\times T(Y),p_1,p_2)$ such that there is a unique morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$ such that $p_1f=T(\pi_1)$ and $p_2f=T(\pi_2)$.

My guess is that this $f$ is the so called natural morphism, but I don't know how to verify that because I don't know what it means. I've only heard of natural transformations/isomorphisms between functors, but not natural morphisms between objects. Can anyone clarify?

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It is common to denote a generic component of a natural transformation by a natural morphism. For example, one says that $V \to V^{**}, v \mapsto (\phi \mapsto \phi(v))$ is a natural linear map, but one actually means a natural transformation $\mathrm{id} \to D \circ D$ where $D$ is the dualization functor. – Martin Brandenburg Nov 17 '13 at 7:31
Thank you @Martin. – Camilla Vaernes Nov 17 '13 at 18:16
up vote 3 down vote accepted

Consider the categories $\mathcal B^2$ and $\mathcal C^2$ of pairs of objects in $\mathcal B$ and $\mathcal C$, respectively. The Cartesian product on $\mathcal B$ is a functor $\times_\mathcal B$ from $\mathcal B^2$ to $\mathcal B$ and similarly for $\mathcal C$. There is also the functor $T^2:\mathcal B^2\to\mathcal C^2$ which applies $T$ to each object in a pair. The "natural morphism" you describe is a natural transformation from the functor $T\circ\times_\mathcal B$ to $\times_\mathcal C\circ T^2$.

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Thank you Smoore. – Camilla Vaernes Nov 17 '13 at 18:17

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