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Let $X$ be a singular projective variety. Denote by $Z$ the singular locus of $X$. Consider the resolution of singularities $$\pi: \widetilde{X} \to X$$ Denote by $E$ the exceptional divisor. We know that there is a long exact sequence in cohomology given by: $$... \to H^{i-1}(E) \to H^i(X) \to H^i(\widetilde{X}) \oplus H^i(Z) \to H^i(E) \to ...$$ Is there a similar exact sequence using Borel-Moore homology, instead of cohomology? If necessary, assume that $Z$ is a union of finitely many points.

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    $\begingroup$ Everything here is compact, so Borel-Moore homology is just homology. $\endgroup$ Commented May 3 at 11:39
  • $\begingroup$ Sure, is there a similar exact sequence using homology? $\endgroup$
    – user45397
    Commented May 3 at 11:57
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    $\begingroup$ Yes: just reverse the direction of the arrows. $\endgroup$ Commented May 3 at 13:23
  • $\begingroup$ Thanks, can you suggest a reference. $\endgroup$
    – user45397
    Commented May 3 at 13:34
  • $\begingroup$ @user45397 Could you give a reference (or a proof sketch) for the above cohomology long exact sequence? $\endgroup$ Commented May 3 at 17:43

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Borel-Moore homology is actually the easiest variant of (co)homology to prove this in. More generally let $\pi \colon Y \to X$ be a proper morphism which induces an isomorphism away from a closed subscheme $Z \subset X$. Write $E := \pi^{-1}(Z)$. Then we have the distinguished triangles $$ C^{BM}_*(Z) \to C^{BM}_*(X) \to C^{BM}_*(X-Z) \to $$ $$ C^{BM}_*(E) \to C^{BM}_*(Y) \to C^{BM}_*(Y-E) \to $$ There is a map of distinguished triangles, from bottom to top, given by proper pushforward, which induces a quasi-isomorphism on the cones/cofibers. So the square $$\require{AMScd}\begin{CD} C^{BM}_*(E) @>>> C^{BM}_*(Y) \\ @V{\pi_*}VV @V{\pi_*}VV \\ C^{BM}_*(Z) @>>> C^{BM}_*(X) \end{CD}$$ is homotopy coCartesian as it induces quasi-isomorphisms on the cones of the horizontal maps. Equivalently there is a distinguished triangle $$ C^{BM}_*(E) \to C^{BM}_*(Y) \oplus C^{BM}_*(Z) \to C^{BM}_*(X) \to $$ which gives rise to the long exact sequence you asked for.

When $X$ is proper, you can then rewrite this using homology or cohomology.

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