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Let $\phi:X\longrightarrow Y$ be a morphism of schemes, and let $y\in Y$. Let $k(y)$ be the constant sheaf $k(y)$ on the closed subset $\{\bar{y}\}$. Then what is $\phi^*(k(y))$?

By definition, $\phi^*(k(y))=\phi^{-1}k(y)\otimes_{\phi^{-1}\mathcal{O}_Y}\mathcal{O}_X$.

So, the stalk $\phi^*(k(y))_x=(\phi^{-1}k(y)\otimes_{\phi^{-1}\mathcal{O}_Y}\mathcal{O}_X)_x$ = $(\phi^{-1}k(y))_x\otimes_{\phi^{-1}\mathcal{O}_{Y_x}}\mathcal{O}_{X,x}$ = $k(y)_{\phi(x)}\otimes_{\mathcal{O}_{Y,\phi(x)}}\mathcal{O}_{X,x}$.

So the stalk, $\phi^*(k(y))_x = 0$ if $\phi(x)\notin \{\bar{y}\}$, and is equal to $k(y)\otimes_{\mathcal{O}_{Y,\phi(x)}}\mathcal{O}_{X,x}$ if $\phi(x)\in \{\bar{y}\}$.

So is this a skyscraper sheaf too? Can we think of this as $k(y)$ is some sense?

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"So is this a skyscraper sheaf too? Can we think of this as $k(y)$ is some sense?"
The answer to both questions is definitely: No!

For example if $k$ is a field, take $X=\mathbb A_k^1$, $Y=\operatorname {Spec}(k)=\{y\}$ and $\phi:\mathbb A_k^1\to \operatorname {Spec}(k)$ the unique $k$-morphism.
Then the structure sheaf $\mathcal O_Y$ is the skyscraper sheaf on $Y$ with stalk $k$ at $y$, but its algebraic inverse image is $\phi^*(\mathcal O_Y)=\mathcal O_X$ (a formula valid for any morphism of schemes $\phi:X\to Y$), which is as far as can be from a skyscraper sheaf and can certainly not be thought of as $k(y)=k$ in any sense.

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