How do we get a simplicial homology functor?

Question

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.

Answer 1

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.

Answer 2

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.

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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.