# Does retracting commute with taking quotients?

Oftentimes I see that when you have a quotient $$Y=X/{\sim}$$ and you want to retract it to some subspace $$B\subset Y,$$ you first retract $$X$$ to $$A \subset X$$ and then observe that $$B=A/{\sim}$$.

I have seen this done a few times now, without explicitly saying it. So I looked up for the actual statement, which I find reasonable, and found it stated and proved without details here.

But if you take $$Y=P^2(\mathbb{R})$$ the real projective space and $$X=D^2=\{x \in \mathbb{R}^2: \|x\|\leq 1\},$$ you can contract $$D^2$$ to an internal point $$P,$$ which would imply by the above reasoning that the real projective space was contractible.

So what am I missing?

• Note my edit to the question. There is a reason why the way I wrote it is how it should be done. (I'm beginning to realize that mathematicians who are unaware of this walk through the same hallways as the rest of us and look like normal humans. Or in other words, it is commonplace and I would not have expected it to be commonplace.) $\qquad$ – Michael Hardy Aug 12 '20 at 17:31
• To be a bit more explicit, look at the conspicuous typographical difference: \begin{align} & X/\sim \\ & X/{\sim} \end{align} The latter is coded as X/{\sim} and the former as X/\sim. This is not a software bug or an unintended quirk. Rather there is good reason to design the software to work that way. Remember that this software ultimately evolved from software written by Donald Knuth, who is not only a genius but is extremely attentive to, and conscientious about, just this sort of thing. – Michael Hardy Aug 12 '20 at 17:37

Let's follow what happens when we take a deformation retraction of $$D^2$$ to an internal point $$p$$, and attempt to use it to induce a deformation retraction of the quotient space $$P^2(\mathbb R)$$. Let's consider the quotient map $$f : D^1 \to P^2(\mathbb R)$$. Under that quotient map, for each $$q \in \partial D^2$$ its opposite boundary point $$-q \in \partial D^2$$ has the same image $$f(q)=f(-q)$$, a point of $$P^2(\mathbb R)$$ that I will denote $$[q]$$. Thus, the quotient map $$f$$ is two-to-one on $$\partial D^2$$, and is otherwise one-to-one.
Now, the function $$f : D^2 \to P^2(\mathbb R)$$ is indeed homotopic to, let's say, the constant map with value $$p = f(0,0)$$: simply use the homotopy $$h((x_1,x_2),t) = f((1-t)x_1,(1-t)x_2)$$.
But if you attempted to use this formula to define a homotopy from $$P^2(\mathbb R)$$, then for each $$q \in \partial D^2$$ you would be faced with an agonizing choice: Does the point $$[q]=[-q] \in P^2(\mathbb R)$$ follow the homotopy path $$f((1-t)q_1,(1-t)q_2)$$? Or does it instead follow the homotopy path $$f((1-t)(-q_1),(1-t)(-q_2))$$?
If these homotopy paths were the same, then there would be no trouble (and that's most likely what is going on the examples you allude to in your opening sentence). BUT, in this example those two paths are not the same. So our attempt to use the given information, i.e. to use the function $$f$$ and the homotopy $$h$$, in a well-defined and continuous fashion to build a deformation retraction from $$P^2(\mathbb R)$$ to a point, has broken down.