What is the induced orientation on a product of vector spaces in singular cohomology? In singular cohomology, we understand an orientation of $\Bbb{R}^d$ as a choice of generator for the relative cohomology group $H^d(\Bbb{R}^d,\Bbb{R}^d\setminus \{0\})\cong \Bbb{Z}$.
My question is whether given orientations $\eta,\xi$ on $\Bbb{R}^d$, $\Bbb{R}^e$, one obtains an orientation on $\Bbb{R}^{d+e}$ by taking the cup product $p_1^\ast(\eta) \smile p_2^\ast(\xi)$, where $p_1:\Bbb{R}^{d+e}\rightarrow \Bbb{R}^d$, $p_2:\Bbb{R}^{d+e}\rightarrow \Bbb{R}^e$ are projections onto the first $d$, last $e$ factors. This makes sense since $p_1^\ast(\eta)$ naturally lives in $H^d(\Bbb{R}^{d+e},(\Bbb{R}^d\setminus \{0\})\times \Bbb{R}^e)$, and $p_2^\ast(\xi)$ naturally lives in $H^e(\Bbb{R}^{d+e}, \Bbb{R}^d \times (\Bbb{R}^e\setminus \{0\}))$, so that their cup product can be considered as lying in $$H^{d+e}(\Bbb{R}^{d+e}, ((\Bbb{R}^d\setminus \{0\})\times \Bbb{R}^e )\cup( \Bbb{R}^d \times (\Bbb{R}^e\setminus \{0\})))=H^d(\Bbb{R}^{d+e},\Bbb{R}^{d+e}\setminus \{0\}).$$
More specifically, I was running into trouble verifying this in the case $d=e=1$.
Indeed, let $[\eta]\in H^1(\Bbb{R},\Bbb{R}\setminus\{0\})$ be given by taking $\eta$ to be the coboundary of the $0$-cochain corresponding to the function
$$ f(x) =
\begin{cases}
0,  & \text{if $x<0$}, \\
1, & \text{if $x\geq 0$}.
\end{cases}
$$
Then if $p_1^\ast([\eta]) \smile p_2^\ast([\eta])$ is to generate $H^2(\Bbb{R}^2,\Bbb{R}^2\setminus \{0\})$, it should pair non-trivially with a generator of $H_2(\Bbb{R}^2,\Bbb{R}^2\setminus \{0\})$, by the Universal Coefficient Theorem.
A generator of this last group is given via the singular $2$-simplex $\sigma:\Delta^2 \rightarrow \Bbb{R}^2$ corresponding to linearly embedding with vertices at $\sigma(v_0)=(-1,2)$, $\sigma(v_1)=(-1,-1)$, $\sigma(v_2)=(2,-1)$. (So that the origin is in the interior of the image of $\sigma$.)
However
\begin{align}
( p_1^\sharp(\eta) \smile p_2^\sharp(\eta) )( \sigma ) &= 
     p_1^\sharp(\eta)( \sigma|[v_0,v_1] ) \cdot p_2^\sharp(\eta)( \sigma|[v_1,v_2] ) \\
&= \eta ( p_1 \circ \sigma|[v_0,v_1] ) \cdot \eta ( p_2 \circ \sigma|[v_1,v_2] ) \\
&= 0,
\end{align}
since $ p_1 \circ \sigma|[v_0,v_1] $ stays in $\Bbb{R}_{<0}$ so gets killed by $\eta$.
Where have I gone wrong?
 A: This is an incredibly annoying subtlety. To unwind, we have to review how cup products work. Let $X$ be a topological space and $A,B\subseteq X$ be subspaces. In its most primitive form, the cup product is a map $C^i(X)\times C^j(X)\rightarrow C^{i+j}(X)$. This restricts to a cup product on relative cochains $C^i(X,A)\times C^j(X,B)\rightarrow C^{i+j}(X,A+B)$. On the other hand, there is an inclusion map $C^{i+j}(X,A\cup B)\rightarrow C^{i+j}(X,A+B)$. Now, all of these maps are compatible with the differential in an appropriate sense, hence induce maps in cohomology $H^i(X,A)\times H^j(X,B)\rightarrow H^{i+j}(X,A+B)$ and $H^{i+j}(X,A\cup B)\rightarrow H^{i+j}(X,A+B)$. The important feature now is that if $A$ and $B$ are open in $A\cup B$, then this latter map is an isomorphism. Composing with its inverse yields the true relative cup product on cohomology, $H^i(X,A)\times H^j(X,B)\rightarrow H^{i+j}(X,A\cup B)$. The important subtlety is that this inverse $H^{i+j}(X,A+B)\rightarrow H^{i+j}(X,A\cup B)$ is not given by the identity on representatives.
If we now apply this discussion to the scenario at hand, you will see that the cocycle $p_1^{\ast}\eta\cup p_2^{\ast}\eta$ represents an element of $H^2(\mathbb{R}^2,\mathbb{R}\times\mathbb{R}\setminus\{0\}+\mathbb{R}\setminus\{0\}\times\mathbb{R})$, but it does not represent an element of $H^2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$ directly. I leave it as an exercise to find an explicit $2$-chain  (a $2$-simplex will do, actually) in $\mathbb{R}^2\setminus\{0\}$ on which $p_1^{\ast}\eta\cup p_2^{\ast}\eta$ does not vanish. Nonetheless, the cohomology class $[p_1^{\ast}\eta\cup p_2^{\ast}\eta]\in H^2(\mathbb{R}^2,\mathbb{R}\times\mathbb{R}\setminus\{0\}+\mathbb{R}\setminus\{0\}\times\mathbb{R})$ defines a cohomology class in $[\eta]\times[\eta]\in H^2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$, which we can pair with the generator $[\sigma]\in H_2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$. The downside is that we don't have an explicit representative of $[\eta]\times[\eta]$, which begs the question how we can actually concretely calculate this evaluation. Let me address this.
I will first go the long route and sketch a geometric way of understanding the inverse $H^{\bullet}(X,A+B)\rightarrow H^{\bullet}(X,A\cup B)$, which will explain how to do this calculation. Then I will give a formal argument. The chain map $C^{\bullet}(X,A\cup B)\rightarrow C^{\bullet}(X,A+B)$ is dual to the quotient map $C_{\bullet}(X,A+B)\rightarrow C_{\bullet}(X,A\cup B)$, which fits into a short exact sequence of chain complexes
$$0\rightarrow C_{\bullet}(A\cup B,A+B)\rightarrow C_{\bullet}(X,A+B)\rightarrow C_{\bullet}(X,A\cup B)\rightarrow0.$$
The complex $C_{\bullet}(A\cup B,A+B)$ is acyclic (this is the excision theorem), whence the other map induces isomorphisms in homology. Geometrically, this acyclicity comes from the fact that a cycle in $A\cup B$ possesses a subdivision, which is a cycle in $A+B$. Similarly, an inverse $H_{\bullet}(X,A\cup B)\rightarrow H_{\bullet}(X,A+B)$ can be obtained by taking a relative cycle in $(X,A\cup B)$, i.e. a chain with boundary in $A\cup B$, and subdivide it (which subdivides the boundary as well, since subdivision is a chain map) until the boundary is a chain with boundary in $A+B$, whence the subdivided cycle represents a homology class in $(X,A+B)$. Lastly, the desired inverse $H^{\bullet}(X,A+B)\rightarrow H^{\bullet}(X,A\cup B)$ is obtained by the dual of this on representatives. This is only a sketch as I'm being intentionally vague about how exactly we should subdivide each chain, etc..
Formally, the point is that the maps $H^{\bullet}(X,A\cup B)\rightarrow H^{\bullet}(X,A+B)$ and $H_{\bullet}(X,A+B)\rightarrow H_{\bullet}(X,A\cup B)$ are adjoints with respect to the pairings $H^{\bullet}(X,A+B)\times H_{\bullet}(X,A+B)\rightarrow\mathbb{Z}$ and $H^{\bullet}(X,A\cup B)\times H_{\bullet}(X,A+B)\rightarrow\mathbb{Z}$ (which is a purely algebraic fact about chain maps and their duals).
Thus, to evaluate $[\eta]\times[\eta]\in H^2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$ on $[\sigma]\in H_2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$, we choose a homology class $\alpha\in H_2(\mathbb{R}^2,\mathbb{R}\times\mathbb{R}\setminus\{0\}+\mathbb{R}\setminus\{0\}\times\mathbb{R})$ that maps to $[\sigma]$ under the canonical map (this can be achieved by appropriately subdividing a representative) and evaluate on it the cocycle $p_1^{\ast}\eta\cup p_2^{\ast}\eta$.
I leave it to you to carry this out explicitly. An appropriate subdivision of $\sigma$ will involve at least four simplices, so might be a bit fidgety. Alternatively, you could consider an appropriate subdivision of the square $[-1,1]^2$ into two simplices, which will yield an easier calculation. Of course, you can also feel free to do both calculations and confirm they yield the same result.
