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I'm trying to understand what a natural transformation is. To this end, I want to show the following:

For each group $H$ the map $G \mapsto H \times G$ defines a functor $H \times -:\textbf{Grp} \to \textbf{Grp}$ and for each group homomorphism $f: H \to K$ there is a natural transformation $H \times - \to K \times -$.

The definition of natural transformation is the following: Let $C,D$ be two categories and $F,G: C \to D$ be two functors. Then a natural transformation $\eta$ is a collection of morphisms $(\eta_A)_{A \in \text{Obj}(C)}$ such that for all objects $A,B \in \text{Obj}(C)$ and all morphisms $\alpha : A \to B$ we have $ \eta_B \circ F(\alpha) = G(\alpha) \circ \eta_A$

Let $f: H \to K$ be a group homomorphism. Then we define $\eta_G$ to be the map $G \times H \to G \times K$, $(g, h) \mapsto (g,f(h)) $ and similarly $\eta_{G^\prime}: G^\prime \times H \to G^\prime \times K$, $(g^\prime, h) \mapsto (g^\prime, f(h))$.

Let $\alpha : G \to G^\prime$ be a group homomorphism and let $(g,h) \in G \times H$. Then $ \eta_{G^\prime} \circ (\times H (\alpha)) (g,h) = \eta_{G^\prime} (\alpha(g), h) = (\alpha(g), f(h))$.

Also, for $(g, h) \in G \times H$ we have $(\times K (\alpha)) \circ \eta_G ((g,h)) = (\times K (\alpha)) ((g, f(h))) = (\alpha(g), f(h))$.

Hence the diagram commutes and we have a natural transformation $\eta : H \times - \to K \times -$.

Can you tell me if this is correct?

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    I think it would be more precise to say '*we have a natural transformation $\eta : - \times H \to - \times K$*' instead, since you "define $\eta_G$ to be the map $G \times H \to G \times K$, $(g, h) \mapsto (g,f(h)) $" and the same goes for $\eta_{G'}$. We can only say that $X \times Y$ is isomorphic to $Y \times X$, but they are not *really* equal.2015-12-30

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This looks correct to me. You have $(g,h)\stackrel{F_H(\alpha)}{\mapsto} (\alpha(g),h)\stackrel{\eta_{\,G\,'}}{\mapsto}(\alpha(g),f(h))$ and $(g,h)\stackrel{\eta_{\,G}}{\mapsto}(g,f(h))\stackrel{F_K(\alpha)}{\mapsto}(\alpha(g),f(h))$ hence $G\times H\xrightarrow{F_H(\alpha)}G'\times H\xrightarrow{\eta_{\,G\,'}}G'\times K$ and $G\times H\xrightarrow{\eta_{\,G}}G\times K\xrightarrow{F_K(\alpha)}G'\times K$ commute (if we were to draw them together in a diagram), i.e. $\eta_{\,G\,'}\circ F_H(\alpha)=F_K(\alpha)\circ\eta_G$, as desired.

(And $F_H(\alpha)$ is $\times H(\alpha)$ and $F_K(\alpha)$ is $\times K(\alpha)$ in your notation.)

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    @magma: typo; fixed.2012-06-10