# $\mathrm{Hom}(\mathcal{O}_p,\mathcal{O}_p[1])=T_pX$?

I'm reading a paper and came across the following isomorphism:

Let $X$ be a $n$-dimensional variety, $p$ a smooth point. Let $\mathcal{O}_p$ be the skyscraper sheaf at $p$.

Then $\mathrm{Hom}(\mathcal{O}_p,\mathcal{O}_p[1])=T_pX$, where [1] indicates the shift ( in the derived category of $\mathrm{Coh}(X)$).

I do not see this. Any pointers? thank you very much!

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I proved this here the other day, math.stackexchange.com/questions/75673/… If the answers in that question solve your problem, we should close this as a duplicate. –  Mariano Suárez-Alvarez Nov 5 '11 at 20:12
Great Mariano, thanks for the link. I suppose Hom(O_p,O_p[n|)=\Lambda^n T_pX is shown similarly? Thanks. –  Carsten Nov 5 '11 at 22:20
You can try and tell us :) –  Mariano Suárez-Alvarez Nov 6 '11 at 1:36
(Also, please do not write titles which are all TeX) –  Mariano Suárez-Alvarez Nov 6 '11 at 2:10
ok, thanks. I will. –  Carsten Nov 6 '11 at 3:50