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.



Abstract simplicial complexes - Reference for an elementary definition of mapping degree for simplicial maps?
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:


1. 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).
2. 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:


1. It bounds two $n$-faces, both of which lie over $\sigma$, contributing opposite signs to the sum defining $\deg(f)$ via $\sigma$.
2. It bounds two $n$-faces, both of which lie over $\sigma'$, contributing opposite signs to the sum defining $\deg(f)$ via $\sigma'$.
3. 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).


