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i want to to the following exercise:

Let $U:\mathcal{C}\rightarrow\mathcal{D}$ be a functor, $F:Obj(\mathcal{D})\rightarrow Obj(\mathcal{C})$ be a function. Assume that there is a $Obj(\mathcal{D})$ indexed family $\{\alpha_X\}$ of $\mathcal{D}$-maps $\alpha_X:X\rightarrow U(F(X))$. Now suppose that for all $f:X\rightarrow U(Y)$ in $\mathcal{D}$ there exists an unique $\tilde{f}:F(X)\rightarrow Y$ in $\mathcal{C}$ such that $U(\tilde{f})\circ\alpha_X=f$ (soory i can not draw good commuting diagramms). Define $F$ on maps to make a functor such that $\{\alpha_X\}$ is a natural transformation.

I have llok to this exercise and thought about the following. Let $g:X\rightarrow Y$ be an arrow in $\mathcal{D}$ then define $F(g)=\tilde{g}$. Is this a good choise and is this choise good enough to end the exercise?

Can someone help me by hints or solution?

Thank you.

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1 Answer 1

For $g:D_1\to D_2$ in $\mathcal D$, extend it to $f:=\alpha_{D_2}\circ g:D_1\to UF(D_2)$. By hypothesis, for this arrow there exists a unique $\tilde f:F(D_1)\to F(D_2)$. So, let $Fg:=(\alpha_{D_2}\circ g)^{\tilde{}}$.

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