How to prove continuity of the retraction map in proof of Brouwer's fixed point theorem? I was reading Allen Hatcher's proof of the Brouwer's fixed point theorem using the no retraction theorem. That is, we want to prove that every continuous $f: \mathbb{D}^n \to \mathbb{D}^n$ has a fixed point $f(x) = x$, by contradiction.

The no retraction theorem says there is no continuous $r: \mathbb{S}^{n-1} \to \mathbb{D}^n$ such that $r \mid_{\mathbb{S}^{n-1}}(y) = y$. i.e. $r$ restricted to $\mathbb{S}^{n-1}$ is the identity functon.

Suppose a continuous $f$ exists with no fixed point. Then, define $r(x)$ by the point on the line $tx + (1-t)f(x)$ intersecting $\mathbb{S}^{n-1}$. The book, and lecture notes that I looked up, say this is clearly continuous.

Question: How do we proof that $r(x)$ is continuous?
You could try the following: Take the function
$$g(t,x) = |tx+(1-t)f(x)|$$ 
and apply the implicit function theorem to get a continuous $t(x)$ with 
$$g(t(x),x) = 1.$$
Then
$$r(x) = t(x) x + (1-t(x))f(x)$$
clearly is continuous since it is constructed from continuous functions.
One way is by showing that : i ) Multiplication is a continuous function  ii)so is sum, i.e., they are both continuous functions $\mathbb R \times \mathbb R \rightarrow \mathbb R$. And iii) The sum of continuous functions is continuous. Then $r(t)$ is a sum of the continuous functions $tx$ and $(1-t)f(x)$. If you want to make sure that the restriction to $S^{n-1}$ is continuous, you can either just say that a continuous function in a space restricts to a continuous function in a subspace, or you can divide $r(t)$ by the (continuous )norm function.