I am interested to use the mapping degree for simplicial maps between (oriented) abstract simplicial complexes. What I mean by "elementary": My preference would be to use a definition of mapping degree that is not based on homology and not based on the geometric realization of the complexes.
As I am new to this area, I have consulted several sources to find a comprehensive definition. In Allen Hatcher's and in Sergey Matveev's Algebraic Topology books it is defined through homology. In Jiri Matousek's book about the Borsuk-Ulam Theorem it is defined for simplicial complexes without orientation, so it only defines degree mod 2. In Wikipedia, nCatLab, and Wolfram MathWorld I could not find it either. In Brouwer's original paper from 1912 the definition of mapping degree is for simplicial complexes, but I find it extremely hard to follow. And if I understand it correctly, it is based on the geometric realization of the simplicial complex.
Hence my question, are you aware of a good source and reference for an "elementary" yet rigorous definition?
EDIT: Based on good comment from Achim Krause below, restricting my request to oriented n-dimensional simplicial complexes all of whose (n−1)-dimensional faces are contained in exactly two n-dimensional ones, with appropriate orientations.
Best Answer
I think the right generality to restrict to is the following:
Let $n$ be a positive integer, and let $X$ and $Y$ be $n$-dimensional oriented simplicial complexes, with the following properties:
Every $n-1$-face is contained in exactly two $n$-faces (with opposite orientations relative to the $n-1$-face, i.e. if $\sigma$ is an $n-1$-face, there should be exactly two elements $x$ such that $\{x\}\cup \sigma$ is an $n$-face, with opposite orientations. Here and below I think of an orientation as assigning to each ordering of the vertices of each $n$-simplex a number $+1$ or $-1$ which flips if the ordering is changed by an odd permutation).
In $Y$, it is possible to go from any $n$-face to any other $n$-face by a path of neighbouring $n$-faces (where two $n$-faces are neighbouring if they share an $n-1$-face). I thought at first this would follow from 1. and connectedness, but something like "Two copies of $\partial\Delta^3$ glued together along a single vertex" shows that this is an independent condition.
Then:
Definition. The mapping degree of a map $f: X\to Y$ is computed as follows: Pick an $n$-simplex $\sigma$ of $Y$, and let $$ \operatorname{deg}(f) := \sum_{\widetilde{\sigma}} \varepsilon_{\widetilde{\sigma}} $$ where the sum is over all $n$-simplices $\widetilde{\sigma}$ in $X$ which are preimages of $\sigma$, and $\varepsilon_{\widetilde{\sigma}} \in \{+1,-1\}$ depending on whether the orientations of $\widetilde{\sigma}$ and $\sigma$ agree or disagree under $f$.
Lemma. The mapping degree is well-defined, i.e. doesn't depend on the choice of $\sigma$ in $Y$.
Proof. Since by assumption we can pass from any $n$-face to any other $n$-face through a sequence of neighbouring $n$-faces, we only need to discuss the case of neighbouring $n$-faces. Let $\sigma$, $\sigma'$ be two neighbouring $n$-faces in $Y$, and let $\tau = \sigma\cap \sigma'$ be their shared $n-1$-simplex. Then every preimage $\widetilde{\tau}$ of $\tau$ in $X$ satisfies exactly one of the following:
It bounds two $n$-faces, both of which lie over $\sigma$, contributing opposite signs to the sum defining $\deg(f)$ via $\sigma$.
It bounds two $n$-faces, both of which lie over $\sigma'$, contributing opposite signs to the sum defining $\deg(f)$ via $\sigma'$.
It bounds two $n$-faces, one lying over $\sigma$, one over $\sigma'$, contributing equal signs to each of the sums defining $\deg(f)$ via $\sigma$ or $\sigma'$.
In all cases we see that the contributions to $\deg(f)$ defined via either $\sigma$ or $\sigma'$ agree. Since every $n$-simplex lying over $\sigma$ or $\sigma'$ appears exactly once in this enumeration, this proves that $\deg(f)$ defined via $\sigma$ or $\sigma'$ agree. $\square$
Note that this is an unwinding of the definition of mapping degree via cohomology (which agrees, by the universal coefficient theorem, with the mapping degree defined via homology). Indeed, under the assumptions discussed above, $H^n(Y)\cong \mathbb{Z}$ is generated by a cocycle taking a fixed positively oriented $n$-simplex to $1$, and all other $n$-simplices to $0$ (and any two such cocycles are cohomologous). To determine the map $f^*: H^n(Y)\to H^n(X)$, we precompose one such cocycle, and decompose the resulting cocycle on $X$ again as sum of such cocycles (which gives the signed sum over preimages).