# Is the homotopy class given by the degree?

Let $X$ be a topological space such that $\pi_n(X)=H_n(X)=Z$. A continuous map $f: S^n \rightarrow X$ is an element of $\pi_N(X)=Z$ therefore $[f]_{\mathrm{homotopy}}$ is characterised by an integer $k$.

I am trying to show that $k$ is the degree of $f$. Is the following correct?

By Hurewicz theorem we have a group homomorphism $h:\pi_n(X)\rightarrow H_n(X)$. Let $[c]$ be a generator of $H_n(X)$, i.e. $[c]_{\mathrm{homology}}=1$. The Hurewicz map is $$h([f]_{\mathrm{homotopy}})=f_*[c]_{\mathrm{homology}} .$$ For $[f]\in\pi_n(X)$ on one hand we have, by definition of degree $$h[f]=f_*[c]=\mathrm{deg}(f)\,[c].$$ On the other hand if $[f]=k$, since the Hurewicz map is a group homomorphism we also have $$h[f]=h[k]=k\, h[1]=k\, [c].$$ It follows that the integer characterising the homotopy class of $f$ is the degree of $f$.

• how do you know that $H_n(X)=\mathbb{Z}$? – Lee Mosher Aug 4 '14 at 18:49
• @LeeMosher You are right, I started with $X=S^n$ and then incorrectly generalised, let me edit the question and put it as an hypothesis. – GFR Aug 4 '14 at 18:52

Here is a counterexample. Consider $X = \mathbb{R}P^3$, an orientable 3-manifold and so $H_3(\mathbb{R}P^3)=\mathbb{Z}$. The universal covering map $S^3 \to \mathbb{R}P^3$ induces an isomorphism on $\pi_3$ (by the long exact sequence of homotopy groups for a covering map). Since the identity map on $S^3$ represents a generator of $\pi_3(S_3)=\mathbb{Z}$ it follows that the universal covering map $S^3 \to \mathbb{R}P^3$ represents the generator of $\pi_3(\mathbb{R}P^3) = \mathbb{Z}$. But the universal covering map has degree 2.
In your proof, the expression $f_*[c]$ makes no sense, because the domain of $f_*$ is $\pi_n(S^n)$ or $H_n(S^n)$ (either one is OK because they are naturally isomorphic by the Hurewicz theorem) but $[c]$ is an element of $H_n(X)$.
• Thinking about what you wrote it seems to me that my "proof" is only valid when there is a map $g: S ^n\rightarrow X^n$ which induces an isomorphism of the $n$th homotopy group AND the Hurewicz map from $\pi_n(X)$ to $H_n(X)$ is an isomorphism because then we have the commutative square $$\begin{CD} \pi_n(S^n) @>g_*>> \pi_n(X)\\@VVhV @VVhV\\H_n(S^n) @>g_*>>H_n(X) \end{CD}$$. Is this correct? I am not strong in algebraic topology, so sorry if I write nonsense and thanks for your help! – GFR Aug 5 '14 at 8:50
• mmm the commutative diagram does not want to compile. The top horizontal line is $\pi_n(S^n) \rightarrow \pi_n(X)$ via the map induced by $g$ on homotopy, the bottom horizontal line is $H_n(S^n)\rightarrow H_n(X)$ via the map induced by $g$ on homology. The vertical maps are the Hurewicz maps. – GFR Aug 5 '14 at 8:57
• Under your assumption about the map $g$, then yes, this is true, and it follows from the theorem that the Hurewicz homomorphism is a natural transformation. – Lee Mosher Aug 5 '14 at 13:03