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We did this proof in class, but there is something I don't understand. So, we constructed the map $q : \mathbb{I} \times \mathbb{S}^{n - 1} \rightarrow \mathbb{B}^n$ defined by $q(t,x) = (1 - t)x$. We proved that this map is quotient map. I agree with the proof at this point. Then, we constructed a map $\phi: \mathbb{I} \times X \rightarrow Y$ defined by the use of the homotopy $F : \mathbb{I} \times \mathbb{S}^{n - 1} \rightarrow Y$. Specifically, it is defined as follows $\phi(x) = F(1 - ||x||, \frac{x}{||x||})$. Alright, now to the part I don't understand. By the univerisal property of quotient map we know that $\phi$ is continous iff $\phi \circ q$ is continous. Why is $\phi \circ q = F(1 - (1 - t)||x||,\frac{x}{||x||})$ continous ?

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    If there is an extension then it is clear? You just use $\phi \circ q$ as homotopy. In the other case you have a zero homotopy, i.e. a map from $\Bbb I \times \Bbb S^{n-1} $ to $Y$ which is constant for say $t=0$. Then universal property tells you that this map factors through $\Bbb B^n$ which is a quotient of $\Bbb I \times \Bbb S^{n-1}$. – 2017-01-14
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    That is the way I would prove it. So I only need that $q$ is a quotient map. Do you know what $X$ is? I don't see what you want to do with it. – 2017-01-14
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    What is $X$? Well, I'm not sure why you want to prove that $\phi\circ q$ is continous when obviously $\phi$ is continous being a (cartesian) product of compositions of continous functions : $F$, $\lVert\cdot\rVert$, $\cdot - \cdot$ and $\frac{\cdot}{\cdot}$. By the same argument $\phi\circ q$ is continous (additionally you have to consider multiplication as a continous function). Unless $0\in X$ for which you would need a special definition for $\phi(0)$. And in that case it is easier to show that $\phi\circ q$ is continous as a composition of continous functions. – 2017-02-20

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