Let $(f,f^{\flat}) \colon (X,\mathcal{O}_X) \to (Y,\mathcal{O}_Y)$ be a morphism of locally ringed space. $f \colon X \to Y$ is a continuous map and $f^{\flat} \colon \mathcal{O}_Y \to f_* \mathcal{O}_X$ is a homomorphism of sheaves of rings.

Assume that $x \in X$. There are $2$ two ways.

[1] The maps $f^{\flat }_U \colon \mathcal{O}_Y(U) \to \mathcal{O}_X(f^{-1}(U))$ for every open neighborhood $U$ of $f(x)$ induce maps $\mathcal{O}_Y(U) \to \mathcal{O}_{X,x}$ and a map $f_x^{\sharp} \colon \mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$

[2] Let $f^{\sharp} \colon f^{-1}\mathcal{O}_Y \to \mathcal{O}_X$ be the morphism corresponding to $f^{\flat}$ by adjointness. Since $(f^{-1}\mathcal{O}_Y)_x = \mathcal{O}_{Y,f(x)}$, we get $f_x^{\flat} \colon \mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ .

I know these $2$ maps coincide. It is possible to represent these maps by taking elements. But I believe there is a proof that is shorter, more suggestive and more category-theoretical.



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