I'm trying to show that there exists a closed, orientable manifold $M^n$ with $\chi(M^n)=k$ for every $k\in\mathbb{Z}$. I was able to show that such a manifold has even dimension using Poincare duality and the universal coefficient theorem.
My attempts at a proof have been constructive so far. I have tried finding a closed, orientable covering space $X$ for $RP^2$ with an "adjustable" number of sheets $k$ (like Hatcher's examples of covering spaces for $S^1\wedge S^1$) so that I would have $\chi(X)=k\chi(RP^2)=k.$
I've also tried using the formulas for the Euler characteristic of products and connected sums $$\chi(M\times N)=\chi(M)\chi(N)$$
$$\chi(M\#N)=\chi(M)+\chi(N)-\chi(S^n)$$
to inductively construct the desired manifold for every $k$.
Any fresh ideas would be greatly appreciated.
Best Answer
You won't get very far by using coverings of $\mathbb{R} P^2$ as only $S^2$ covers $\mathbb{R} P^2$ non-trivially. Instead you can use the following: The $n$-torus $T^n$ has vanishing Euler characteristic, so taking connected sum reduces the Euler characteristic by $2$ if $n$ is even. Furthermore $\chi(\mathbb{C}P^n)=n+1$, hence we can reach every integer by taking the connected sum of the form $\mathbb{C} P^n\# T^{2n}\#\dots\# T^{2n}$.