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Let $X$ be a smooth projective variety over complex numbers. Let $f:Y\hookrightarrow X$ be a smooth closed subvariety. Suppose $i:Z\hookrightarrow X$ be a closed subscheme of $X$ and $j:Z'=Y\times_X Z\hookrightarrow Y$ be the corresponding subscheme of $Y$.

Consider the exact sequence on $X$: $$0\rightarrow I_Z\rightarrow O_X\rightarrow i_*O_Z\rightarrow 0\,.$$

Pull this back by $f$ to $gY$, we get (with kernel $K$): $$0\rightarrow K\rightarrow f^*I_Z\rightarrow O_Y\rightarrow f^*i_*O_Z\rightarrow 0\,.$$

1) Is $f^*i_*O_Z=O_{Z'}$? 2) Also suppose the ideal sheaf $I_{Z'}$ is the image of $f^*I_Z$ in $O_Y$ i.e. $$0\rightarrow K\rightarrow f^*I_Z\rightarrow I_{Z'}\rightarrow 0\,.$$

Then what do we know about $f^*I_Z$ and $K$? What are their ranks? Will $f^*I_Z$ and $I_{Z'}$ be equal on an open set?

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Your 1) is correct, essentially as a consequence of the right exactness of tensor product, which makes the assumption in 2) correct. The latter part, the answer can vary. For example, just to get a feel for the last question (the answer to it in general is no), try working out the case $Y=Z$.

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  • $\begingroup$ when $Y=Z$, we get $0\rightarrow K\rightarrow f^*I_Y\rightarrow O_Y\rightarrow O_Y\rightarrow 0$. In this case the the map $f^*I_Y\rightarrow O_Y$ is the zero map? Can we say what $K\simeq f^* I_Y$ is? $\endgroup$ – user52991 Apr 4 '17 at 16:11
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    $\begingroup$ In this case, $f^*I_Y$ is called the conormal bundle, $I_Y/I_Y^2$. $\endgroup$ – Mohan Apr 4 '17 at 16:47

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