cup product well definedness So the cup product is not well defined over co-chain groups, but all the books claim it is well defined over co-homology groups. The only thing I am not clear on is invariance under ordering/re-ordering of simplices when we go to the co-homology level. Every book seems to gloss over this, and after doing a few examples, I can't seem to figure out how to get this to work out right. Can someone fill me in?
 A: Strictly speaking, the cup product is not commutative, though it is commutative up to sign on the level of cohomology.
There is an abstract way of seeing this: namely, we can use the method of acyclic models.  Consider the following two functors from the category of spaces to the category of chain complexes. The first is $X \mapsto C_*(X \times X)$; the second is  $X \mapsto C_*(X)\otimes C_*(X)$. Since these are free and acyclic functors on the subset of standard simplices (this means that a) both can be represented as a sum of free abelian groups on sets which are representable functors in $X$, represented by simplices and b) evaluated on a simplex, they lead to acyclic complexes), there is a natural chain equivalence between two
$$ C_*(X \times X) \simeq C_*(X) \otimes C_*(X)$$
which itself is unique up to chain homotopy. This is the acyclic model theorem (as in Spanier, for instance).
Now the category of chain complexes over a commutative ring is not just an abelian category; it is a monoidal category. We can tensor two chain complexes and get a new chain complex. Moreover, it is a symmetric monoidal category because there is an isomorphism $A_* \otimes B_* \simeq B_* \otimes A_*$ for chain complexes $A, B$. 
Thus if we are given one chain equivalence (fixed throughout the following) $C_*(X \times X) \simeq C_*(X) \otimes C_*(X)$, we get another by composing it with the swap map on the latter. These are both natural in $X$ and so, by the uniqueness (up to chain homotopy) in the acyclic model theorem, we find that the two maps 
$$C_*(X \times X) \rightrightarrows C_*(X) \otimes C_*(X)$$ are naturally chain homotopic.
But the dual of this means that the two maps
$$C^*(X) \otimes C^*(X) \rightrightarrows C^*(X \times X)$$ are naturally chain homotopic. On the level of cohomology, the two maps $H^*(X) \otimes H^*(X) \rightrightarrows H^*(X \times X)$$ are thus equal.
Now the map $C_*(X \times X) \to C_*(X) \otimes C_*(X)$ is precisely the homology cross product, and its dual is the cohomology cross product. So we have seen that if we consider the cross-product map $H^*(X) \otimes H^*(X) \to H^*(X \times X)$, it is invariant under switching the two factors.
You might now object that I said that the cross-product (and thus the cup-product, which is obtained from the cross product by pulling back by the diagonal) is skew-commutative, not commutative. This comes from a feature of how the tensor product of complexes is actually defined: as a result, when you define the swap morphism, you have to introduce a sign (for it to be a chain map).
A: right, ok. I asked about if cup product is well defined, here is why I am confused.
You wrote:

Strictly speaking, the cup product is not commutative, though it is commutative up to sign on the level of cohomology.
There is an abstract way of seeing this...

The cup product being commutative should have nothing to do with it being well defined.
Commutativity of the the product should be talking about if $a \smile b = b \smile a$. This seems irrelevant for a conversation about the cup product being well defined.
So again, what I want to know if why is the cup product well defined. What I am asking is: is it the case that $(\alpha \smile \beta)(a)$ is the same as $(\alpha \smile \beta)(b)$ whenever $a$ is equivalent to $b$. Where equivalent is as as co-homology classes.
Now in your second comment, you have said that that it is not the case that that the cup product is well defined, and it's again unclear if you mean over co-chain groups or co-homology groups. I
n either case, what I want to know is:
When is the cup product well defined? Does your definition of well defined differ from the classical one? I do not at this point want to underst
A: At the cochain level, for $\varphi \in C^k(X)$ and $\psi \in C^l(X)$, the cup product is defined as, $$(\varphi \smile \psi)(\sigma) = \varphi (\sigma \big|_{\leq k}) \psi (\sigma \big|_{\geq k})$$ and extended linearly, where $\sigma \big|_{\leq k}$ is the restriction to vertices from $1$ to $k$ and $\sigma \big|_{\geq k}$ is the restriction to vertices from $k$ to $k +l$.
It can be shown, by explicit computation that $$ d(\varphi \smile \psi) = d\varphi \smile \psi + (-1)^k \varphi \smile d \psi$$
Now, suppose $[\varphi] \in H^k(X)$ and $[\psi] \in H^l(X)$, where $\varphi , \psi \in \ker d$.
The above equation shows us that $\varphi \smile \psi \in \ker d$.  Also, if $\varphi \in \text{im } d$, ie, if $\varphi = d \varphi '$ for some $\varphi '$, then
\begin{align} 
d(\varphi ' \smile \psi) &= d\varphi '\smile \psi + (-1)^k \varphi '\smile d \psi \\ &=  d\varphi '\smile \psi
\end{align}
as $\psi \in \ker d$;  showing us that $d\varphi ' \smile \psi \in \text{im } d$.
Thus, changing $\varphi$ by an element of $\text{im} d$ does not change the cohomology class of the cup product. Similarly, changing $\psi$ by an element of $\text{im } d$ does not change the cohomology class of $\varphi \smile \psi$ either, showing us that the product is well defined on cohomology.
