If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective. If $f:D^2 \to D^2$ is continuous and $f|S^1=id_{S^1}$, then $f$ is surjective, where $D^2$ is the unit closed disk in $R^2$. 
Is this statement correct? I wrote this down in my note as a corollary to the Brower's fixed point theorem and the no retraction theorem from $D^2$ onto $S^1$, however, I can't figure out why this is true. I'm concerned whether I wrote down something wrong from the lecture. I would greatly appreicate any help.
 A: 
Suppose $f:D^2\to D^2$ has $f|_{\partial D^2}$ mapping to $\partial D^2$. If $f$ is not surjective, then its restriction to $\partial D^2$ is nullhomotopic.

Let $f:D^2\to D^2$ be an extension of $\hat{f}:S^1\to S^1$. If $f(D^2)$ misses a point $p$, then $f$ is homotopic to a map $\tilde{f}:D^2\to S^1$ by composing it with the deformation retraction from $D^2-\{p\}$ to $S^1$ In particular $\tilde{f}|_{\partial D^2} = \hat{f}$, and so $\tilde{f}$ provides a hullhomotopy of $\hat{f}$.

To elaborate on that last sentence, notice that if we parametrize $D^2$ with polar coordinates $(r,\theta)$, $r\leq 1$, then $\tilde{f}(1,\theta) = \hat{f}(\theta)$ and the map $H(t,\theta) = \tilde{f}(t,\theta)$ provides a homotopy in $S^1$ between $\hat{f}$ and a constant map.
Now the desired result follows from the contrapositive of the block quote above.
A: Because $f$ is continuous and $D^2$ is arc connect, $f(D^2)$ is arc connect. Now, because $S^1\in Im(f)$, if $f$ isn't surjective, the image isn't arc connect.
