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:

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

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.

Proof

Fill this in later