TITLE: Different definitions of projective objects
QUESTION [9 upvotes]: There are various characterizations for an $R$-module to be projective. Two of them can be generalized to any category:
i) $P$ is an object such that given morphisms $\alpha: A \rightarrow B$ and $\varphi: P \rightarrow B$ such that $\alpha$ is an epimorphism, there exists a morphism $\widetilde{\varphi}: P \rightarrow A$ such that $\alpha \circ \widetilde{\varphi} = \varphi$. (Lifting property) In other words, the functor $\text{Hom}_{\mathsf{C}}(P,-): \mathsf{C} \rightarrow \mathsf{Set}$ preserves epimorphisms.
ii) $P$ is an object such that every epimorphism with codomain $P$ splits.
It is easy to see that (i) $\Rightarrow$ (ii) holds in general, by taking $\varphi = \text{id}_P$. Also if the category has the property that for every object $X$ there is an epimorphism $\pi:P \rightarrow X$ where $P$ satisfies (i) (an abelian category with this property is said to have  enough projectives), we can show the reverse implication (ii) $\Rightarrow$ (i).
My first question: Does (ii) $\Rightarrow$ (i) hold in general? I suspect the answer is no. If so, under which conditions does it hold?

REPLY [11 votes]: For clarity, let us reserve the phrase projective object for an object satisfying (i) and use choice object for an object satisfying (ii). (One formulation of the axiom of choice is to say that every object in $\textbf{Set}$ is a choice object.)
Proposition. If $\mathcal{C}$ is a category such that

all pullbacks exist, and
epimorphisms are preserved by pullbacks,

then choice objects in $\mathcal{C}$ are projective objects.
Proof. Let $C$ be a choice object in $\mathcal{C}$, let $f : A \twoheadrightarrow B$ be an epimorphism, and let $h : C \to B$ be any morphism. Let $r : A \times_B C \to C$ be the pullback of $f$ along $h$. By hypothesis, $r$ is an epimorphism, so there is a morphism $s : C \to A \times_B C$ with $r \circ s = \textrm{id}$. Let $\ell : A \times_B C \to A$ be the pullback of $h$ along $f$, and set $g = \ell \circ s$. By construction, $$f \circ g = f \circ \ell \circ s = h \circ r \circ s = h$$
so we have the required lifting of $h$ along $f$. Thus $C$ is projective. $\quad \blacksquare$
This is a much milder assumption than assuming $\mathcal{C}$ has enough projectives. For example, every topos and every abelian category satisfies the hypotheses of the proposition. If $\mathcal{C}$ is a regular category then the same proof shows that every choice object is regular-projective. (A regular-projective object is an object with the lifting property for regular epimorphisms.) In some cases every epimorphism is regular: for example, the category of groups is a regular category in which every epimorphism is regular.


TITLE: Prove that the normal to the parabola at $Q$ bisects the angle $FQP$
QUESTION [0 upvotes]: QUESTION: Consider the parabola $y^2=4x$. Let $P=(a,b)$ be any point inside the parabola, i.e. $b^2<4a$ and let $F$ be the focus of the parabola. Find the point $Q$ on the parabola such that $FQ+QP$ is minimum. Also show that the normal to the parabola at $Q$ bisects the angle $FQP$.

MY ANSWER: The first part of the problem is easy. I could do that. The answer to that came out to be $|a+1|$. But I am stuck at the second part. I cannot prove that the normal bisects that angle. Can anyone please help me out?
Thank you.

REPLY [1 votes]: We take parabola equation as
$$ y^2= 4 f x \tag1 $$
Bringing in $f$ instead of unity  has the advantage of checking dimensions of equations and also assigning physical meaning to focal length of parabola $ f=OF $
Differentiate (1)
$$ 2 y y'= 4 f\rightarrow y'= \tan \phi = \dfrac{2f}{\sqrt{4fx}}=\sqrt{f/x} \tag2 $$
From the diagram we need to prove that

$$ \beta + \phi = \pi/2 -\phi  $$
or
$$ \beta  = \pi/2 -2 \phi \tag3 $$
or
$$ \tan \beta = \tan ( \pi/2-2 \phi) \tag4$$
LHS
$$\tan \beta =\dfrac{x-f}{y}=\dfrac{x-f}{2 \sqrt{fx}}=\dfrac{1-f/x}{2\sqrt{f/x}} \tag5 $$
RHS
$$ \cot 2\phi = \dfrac{1- \tan^2\phi}{2 \tan \phi} =\dfrac{1-f/x}{2\sqrt{f/x}} \tag6$$
which tally, the normal at $Q$ reflects incident rays to focus $F$ and vice-versa.