Surjectivity of a map from a space to itself I am wondering how to prove that a non-zero degree map from $A \to A$ is surjective. For example, identifying $S^1 \subset \mathbb{C}$, we can take $f:S^1 \to S^1$ via $f(z) = z^k$ with $k\neq 0$. This is a map of degree $k$. How can I show that it is surjective?
Note that I am interested in general topological spaces, not the circle example in specific.
 A: $f : S^n \to S^n$ be a map of nonzero degree which is not surjective. Then there is a point $x \in S^n$ such that $x$ doesn't lie in the image of $f$. Thus, image of $f$ lies entirely inside $S^n - \{x\} \cong \Bbb R^n$. Now nullhomotope $f$ by straightline homotopy. As $f$ is nullhomotopic, $\text{deg} f = 0$, contradiction.
Hence, $f$ must be surjective.
In general, if $f : M \to N$ is a map of nonzero degree between compact closed orientable $n$-manifolds, then $f$ has to be surjective, as otherwise there is some $x \in N$ such that image of $f$ lies entirely withing $N - \{x\}$. By Poincare duality, $H_n(N - \{x\})$ is isomorphic to $0$-th compactly supported cohomology group $H^0_c(N - \{x\})$, which is the trivial group as $N - \{x\}$ is not compact. Thus, as $H_nf : H_n(M) \to H_n(N)$ factors as $H_n(M) \to H_n(N - \{x\}) \to H_n(N)$, $\deg \, f = 0$, which gives us the desired contradiction.
A: If $f$ is not surjective it has degree zero. To see this, assume there is some $y$ not in the image of $f$, and factor $f:S^1\to S^1$ as $f=h\circ g$, with
$g:S^1\to S^1/\{y\}$, and $h:S^1 / \{y\}\to S^1$. Since $S^1/\{y\}$ is contractible, it's first homology group is the trivial group and so the induced homomorphism $h_{*}=g_{*}h_{*}=0$, hence the degree is zero.
A similar result holds for higher $n$. 
