What is this value called in algebraic topology?

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I have a weak familiarity with homotopies and homologies and I have been thinking about similar constructs on my own in the past. Now I would like to classify these ideas of mine before studying these theories properly, which I plan to do soon. For that sake, I would really appreciate if someone would tell me the name of this invariant, if it exists.

Definition of my value: (not as long as it seems)
Let $f:\mathbb{S}^n\rightarrow\mathbb{R}^{n+1}\setminus\{0\}$ be a continuous function. The value of $f$ which I am talking about is defined similarly to Riemann's integrals. We define a quantization of $f$ of accuracy $\varepsilon$, define the wanted value for these quantizations and, finally, say that the value of $f$ is the limit of values of arbitrary array of quantizations of $f$ whose accuracies are approaching $0$. For example, if $f_n$ is chosen to be any quantization of $f$ of accuracy $\frac{1}{n}$, then the limit of values of $f_n$ is defined to be the value of $f$. It can be shown, just like with the Riemann's integral, that this definition doesn't depend on the array of quantizations itself.

Definition of a quantization:
Start with a continuous function $f:\mathbb{S}^n\rightarrow\mathbb{R}^{n+1}\setminus\{0\}$. Choose the accuracy $\varepsilon<1$. Next, break the domain $\mathbb{S}^n$ into many little simplexes, while taking care that the distance of any two vertices of any simplex is less than $\varepsilon$ (for this, use spherical geometry: a straight line is a part of a great circle, or equivalently, a simplex is any stereographic projection of a simplex; distances are angular). Finally, the set of these simplexes and the restriction of $f$ to the vertices of these simplexes is called a quantization of $f$ of accuracy $\varepsilon$.

Definiton of the value of a quantization:
For this we add up the values of individual simplexes. Take a simplex $S$, with vertices $v_1,\ldots,v_{n+1}$, and the normalized $n+1$ values of $f$ in these vertices: $f_i:=\frac{f(v_i)}{||f(v_i)||}$. Now calculate the oriented volume of the simplex $(f_1,\ldots,f_{n+1})$ and multiply it with the orientation of the original simplex $S$ (for this, we again pull the orientation back from a plane using a stereographic projection; volumes are spherical): this is the value of the simplex $S$. This definition is possible if $||x-y||<\varepsilon\Rightarrow||f(x)-f(y)||<1$, which is satisfied for sufficiently small $\varepsilon$.

Some intuitive interpretation of this and its uses:
It turns out, if I am not wrong, that this definition is valid and that all the values are whole numbers. It implies that the value is invariant under homotopy. It is a generalization of a winding number of a closed curve in a punctured plane. If $f$ was smooth (at least $\mathcal{C}^1$), then this value could be defined using some integral (but I am not literate enough to write that down, seriously). This value, if everything is correct, could be used to prove that two euclidean spaces of different dimension are not homeomorphic: suppose $h:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1+k}$, $k\geq1$ is a homemorphism which maps $0$ to $0$. Take the identity map $g:\mathbb{S}^n\rightarrow\mathbb{R}^{n+1}$. The value of $g$ is $1$. Observe $\gamma:=h\circ g$. "Homotopically" morph this $\gamma$ to some smooth approximation $\gamma'$ of its own which is close enough to $\gamma$ so that it too avoids $0$. Define $\alpha:\mathbb{S}^n\times\mathbb{R}\rightarrow\mathbb{R}^{n+1+k}:(x,y)\mapsto\gamma'(x)\cdot y$, it is a smooth map from a manifold of dimension $n+1$ to a manifold of a greater dimension $n+1+k$, so it has measure $0$. This implies that there exists an affine line through $0$ in $\mathbb{R}^{n+1+k}$ which avoids $\gamma'$. We can now homothetically shrink $\gamma'$ to a non-zero point of that affine line. If we pull these homotopies back to $\mathbb{R}^{n+1}$ using $h^{-1}$, we arrive at a homotopy of $g$ which transforms $g$ into a constant function. The value of any constant function is $0$, so it violates the invariance of the value because it was $1$ for $g$.