Hurewicz map

Definition

Explicit definition

Let $X$ be a path-connected space. For $n$ a positive integer, the $n^{th}$ Hurewicz map based at $x_{0}$ of $X$ is a 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 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.

To note that this gives a well-defined map on $\pi _{n}(X,x_{0})$, 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. Further information: Hurewicz map is well-defined

Hands-off definition

Here is an alternative description of the map. We use the fact that $f:S^{n}\to X$ induces a map between $H_{n}(S^{n})$ and $H_{n}(X)$. But $H_{n}(S^{n})=\mathbb {Z}$ and we can thus simply look at the image of the generator of this, to give an element in $H_{n}(X)$.

Facts

The image of the Hurewicz map

The image of the Hurewicz map is a subgroup comprising those singular homology classes that are represented by a singular simplex with the property that all points of the boundary get mapped to the basepoint $x_{0}$.

In particular, the Hurewicz map being surjective means that every continuous map from $\Delta ^{n}$ to $X$ (and in fact, every formal sum of such continuous maps) is homologous to a continuous map with the property that the entire boundary $\partial \Delta ^{n}$ is mapped to $x_{0}$.

The kernel of the Hurewicz map

The kernel of the Hurewicz map comprises those homotopy classes of maps from $S^{n}$ to $(X,x_{0})$ that are nullhomologous. In the case $n=1$, the explanation lies in non-commutativity, i.e., by cutting and rearranging the pieces of the map, we can get a nullhomotopic map.

Related facts

• Hurewicz theorem: This states that if $X$ is $(n-1)$-connected, then the $n^{th}$ Hurewicz map is an isomorphism (if $n\geq 2$) and is the map to the abelianization (if $n=1$).
• Freudenthal suspension theorem