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.

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 $n$. 
