The following should be something well?-known, but i haven't seen it anywhere, neither have i met any references about.

Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$ colors: $v^{o}\rightarrow w(v^{o})\in$ {1,2,...,n+2} $1,2,...,n+2$ } and we shall say that the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color $i\in$ {1,2,...,n+2} $1,2,...,n+2$ } and consider the $n$-simplices whose vertices are colored with exactly the colors {1,2,...,n+2}$\backslash{i}$. $1,2,...,n+2$ }$\backslash{i}$. Let $\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered according to the positive orientation of $\Delta^{n}$ induced by the orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta ^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of $\Delta^{n}$ is colored $i$, or there are two identically colored vertices. Let finally

$\sigma_{i}(w)=\sum\sigma_{i}(\Delta^{n})$,

where the sum is over all $n$-simplices.

The Claim: The number $\sigma_{i}(w)$ does not depend on $i$: $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant $\sigma(w)$ of the coloring $w$.

This invariant has a geometrical meaning: Consider the dual cell complex of the triangulation $\tau$, then since each cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we get a covering $\lambda=${${F_{1},...,F_{n+2}}$} of $M^{n}$. It is easy to see that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$ into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the $n$-sphere $\mathbb{S}^{n}$: $\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals $\sigma(w)$:

$\deg\varphi=\sigma(w)$.

As the proofs are not sophisticated at all and the construction seems conceptual, maybe it is worth including this material in an elementary topology textbook. Note also that it gives a method for calculating the degree without smooth approximation.

Of course, i don't want to repeat well-known things without citation, so any references are welcome.

The following should be something well?-known, but i haven't seen it anywhere, neither have i met any references about.

Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$ colors: $v^{o}\rightarrow w(v^{o})\in{1,2,...,n+2}$ w(v^{o})\in$ {1,2,...,n+2} and we shall say that the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color $i\in i\in$ {1,2,...,n+2}$ 1,2,...,n+2} and consider the $n$-simplices whose vertices are colored with exactly the colors ${1,2,...,n+2}\backslash{i}$. {1,2,...,n+2}$\backslash{i}$. Let $\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered according to the positive orientation of $\Delta^{n}$ induced by the orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta ^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of $\Delta^{n}$ is colored $i$, or there are two identically colored vertices. Let finally

$\sigma_{i}(w)=\sum\sigma_{i}(\Delta^{n})$,

where the sum is over all $n$-simplices.

The Claim: The number $\sigma_{i}(w)$ does not depend on $i$: $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant $\sigma(w)$ of the coloring $w$.

This invariant has a geometrical meaning: Consider the dual cell complex of the triangulation $\tau$, then since each cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we get a covering $\lambda={F_{1},...,F_{n+2}}$ \lambda=${${F_{1},...,F_{n+2}}$} of $M^{n}$. It is easy to see that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$ into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the $n$-sphere $\mathbb{S}^{n}$: $\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals $\sigma(w)$:

$\deg\varphi=\sigma(w)$.

As the proofs are not sophisticated at all and the construction seems conceptual, maybe it is worth including this material in an elementary topology textbook. Note also that it gives a method for calculating the degree without smooth approximation.

Of course, i don't want to repeat well-known things without citation, so any references are welcome.

The following should be something well?-known, but i haven't seen it anywhere, neither have i met any references about.

Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$ colors: $v^{o}\rightarrow w(v^{o})\in{1,2,...,n+2}$ and we shall say that the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color $i\in {1,2,...,n+2}$ and consider the $n$-simplices whose vertices are colored with exactly the colors ${1,2,...,n+2}\backslash{i}$. Let $\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered according to the positive orientation of $\Delta^{n}$ induced by the orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta ^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of $\Delta^{n}$ is colored $i$, or there are two identically colored vertices. Let finally

$\sigma_{i}(w)=\sum\sigma_{i}(\Delta^{n})$,

where the sum is over all $n$-simplices.

The Claim: The number $\sigma_{i}(w)$ does not depend on $i$: $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant $\sigma(w)$ of the coloring $w$.

This invariant has a geometrical meaning: Consider the dual cell complex of the triangulation $\tau$, then since each cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we get a covering $\lambda={F_{1},...,F_{n+2}}$ of $M^{n}$. It is easy to see that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$ into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the $n$-sphere $\mathbb{S}^{n}$: $\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals $\sigma(w)$:

$\deg\varphi=\sigma(w)$.

As the proofs are not sophisticated at all and the construction seems conceptual, maybe it is worth including this material in an elementary topology textbook. Note also that it gives a method for calculating the degree without smooth approximation.

Of course, i don't want to repeat well-known things without citation, so any references are welcome.