Hurewicz map is well-defined

Loose statement
Let $$X$$ be a path-connected space. For $$n$$ a positive integer, we want to show that the $$n^{th}$$ Hurewicz map based at $$x_0$$ of $$X$$ is a well-defined map:

$$\pi_n(X,x_0) \to H_n(X)$$

where $$\pi_n(X,x_0)$$ is the $$n^{th}$$ homotopy group, and $$H_n(X)$$ is the $$n^{th}$$ singular homology group.

The strict map
The map is defined as follows. First define a map:

$$\eta:\Delta^n \to S^n$$

which essentially uses the identification of $$S^n$$ with the quotient of $$\Delta^n$$ by the collapse of its boundary to a single point, i.e., a homeomorphism $$\Delta^n/\partial \Delta^n \to S^n$$.

Now given any based continuous map $$f: (S^n,*) \to (X,x_0)$$, consider $$f \circ \eta$$. This gives a $$n$$-singular chain in $$X$$, and its homology class is precisely the element we are looking for.

What we need to show
To note that this is indeed well-defined, we need to show that if $$f_1$$ and $$f_2$$ are homotopic maps as based continuous maps from $$(S^n,*)$$ to $$(X,x_0)$$, then $$f_1 \circ \eta$$ and $$f_2 \circ \eta$$ are both in the same homology class.