How do we get a simplicial homology functor? The $n$-th simplicial homology group $H_n(A)$ of an abstract simplicial complex $A$ depends on the choice of an orientation for $A$ (but for different orientations, the homology groups are isomorphic!).
Does that mean that in order to get a functor $$H_n \;\colon\; \mathrm{Smp} \to \mathrm{Ab}$$ I have to choose an orientation for every possible abstract simplicial complex? If so, this will require a version of the axiom of choice for classes. Is this reasonable or is there a nicer way to get such a functor?
Edit (after first few answers):
Many people have proposed to include the orientation of a simplex in its definition (oriented simplex). This is a valid approch of course.
But I should have mentioned that I would like to use, if possible, the (minimalistic) definition of an abstract simplex, which you can find e.g. on Wikipedia. Sorry for not making this clear.
 A: You can define a "simplex" as having an orientation, thus getting an easier answer.
A $k$-simplex is the convex hull of a set of $k+1$ points. But what does it mean for a set to have $k+1$ points? That there is a bijection from $\{1,2,\dots,k+1\}$. So simply define "$k$-simplex" in terms of a map $\{1,2,\dots,k+1\}\to\mathbb R^n$ and you can pick an orientation without choice.

Responding to the change to "abstract simplicial complex." You can still redefine an "abstract simplicial complex" to require that the underlying set is totally ordered, which gives you a workaround.
But there is clearly a problem with the usual definition of the functor without an obvious ordering.
A: The trick is to embed the category of abstract simplicial complexes inside the category of symmetric simplicial sets (= functor $\mathbf{F}^\mathrm{op} \to \mathbf{Set}$, where $\mathbf{F}$ is the category of positive finite cardinals): this can be done by sending an abstract simplicial complex $X$ to the symmetric simplicial set $\mathrm{Hom}(\Delta^{\bullet}, X)$.
Now let $\mathbf{\Delta}$ be the category of positive finite ordinals (and monotone maps). There is an evident embedding $\mathbf{\Delta} \to \mathbf{F}$, so by restriction, every symmetric simplicial set is also a simplicial set (= functor $\mathbf{\Delta}^\mathrm{op} \to \mathbf{Set}$). It is straightforward to define a homology functor $H_* : [\mathbf{\Delta}^\mathrm{op}, \mathbf{Set}] \to \mathbf{Gr Ab}$.
Of course, what one has to show is that putting all this together recovers the traditional definition of simplicial homology for an abstract simplicial complex. Let $\Delta^n$ be the standard $n$-simplex (with its canonical ordering), let $X$ be an abstract simplicial complex, let $Y$ be the corresponding symmetric simplicial set, and choose a linear ordering of the vertices of $X$. Then the chain complexes $C_{\bullet} (X)$ and $C_{\bullet} (Y)$ are defined as follows:


*

*$C_{n} (X)$ is freely generated by the set of (non-degenerate) $n$-simplices of $X$, and the boundary map $C_{n} (X) \to C_{n-1} (X)$ is defined by the usual alternating sum.

*$C_{n} (Y)$ is freely generated by the set of (possibly degenerate) simplicial maps $\Delta^n \to X$, and the boundary map $C_{n} (Y) \to C_{n-1} (Y)$ is defined by the alternating sum where the numbering is induced by the canonical ordering of $\Delta^n$ (not the linear order of the vertices of $X$).


The linear ordering of the vertices of $X$ induces a chain map $C_{\bullet} (X) \to C_{\bullet} (Y)$ that sends each (non-degenerate) $n$-simplex of $X$ to the unique order-preserving simplicial map $\Delta^n \to X$ whose image is that $n$-simplex. I claim that the induced homomorphism $H_* (X) \to H_* (Y)$ is an isomorphism. I do not personally know an elementary proof of this, but in principle it should be possible to prove directly that $C_{\bullet} (X) \to C_{\bullet} (Y)$ is a chain homotopy equivalence.
