Hurewicz map is well-defined

Statement

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:

$\eta :{\mathrm{\Delta }}^n\to S^n$

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

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.

Proof

Fill this in later