Orientability and blow-ups

Question

Is it true that if I have a smooth (if you want algebraic) orientable surface (as a real manifold) then blowing up at a point will yield a non-orientable surface and vice versa (i.e. blowing up a non-orientable surface on a point will yield an orientable one)? If so, what is the reason or a rough proof of this fact (or at least a reference to this)? If not, what would be a nice counterexample?

Answer 1

There's a handy short exact sequence for the cohomology of a blow-up which follows from the proper base change theorem (I believe this is covered in Griffiths and Harris); this can be used together with the cohomological description of orientability: if $X$ is a real surface, $p \in X$ a point, $Y = \text{Bl}_p X \to X$ the blowup at $p$ of $X$ and $E \subset Y$ the fibre of the point $p$, then there's a long exact sequence in singular cohomology groups $$ H^1(E,\mathbf{Z}) \to H^2(X,\mathbf{Z}) \to H^2(Y,\mathbf{Z})\oplus H^2(\{p\},\mathbf{Z})=H^2(Y,\mathbf{Z}) \to H^2(E,\mathbf{Z})=0 $$ where on the right we have the vanishing of $E$'s second singular cohomology since it's a curve and thus has vanishing cohomology in degrees higher than $1$.

From this we see that $X$'s top cohomology surjects onto that of $Y$, with a kernel given by (the image of) $E$'s top cohomology. This implies that $Y$ is never orientable if $X$ isn't, and if $X$ is orientable then one can use that (under mild assumptions on $X$, satisfied for instance if $X$ is smooth) $E$ is isomorphic to the real projective line $E \cong \mathbf{R}P^1 \cong S^1$ whose top cohomology is given by $\mathbf{Z}$. Hence $H^2(Y,\mathbf{Z})$ is the non-zero cokernel of a non vanishing map $\mathbf{Z} \to \mathbf{Z}$ which is always torsion $\implies Y$ is never orientable.

I hope I didn't mess something up since I'm not quite used to real manifolds :P let me know if this helps! :)

edit: these arguments are only valid under the additional assumption that $X$ - and hence $Y$ - is compact.