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 Post subject: Category Theory for beginners (2)Posted: Sat Jan 16, 2016 11:04 pm
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 312
(I) Let $\displaystyle f : A \longrightarrow B$ be a morphism in a category $\mathcal{C}$. Show that:

1. If $\displaystyle f$ is a section, then $\displaystyle f$ is monic.
2. If $\displaystyle f$ is a retraction, then $\displaystyle f$ is epic.

(II) If $\displaystyle f$ is a morphism in a category $\mathcal{C}$, then prove that the following are equivalent:

1. $\displaystyle f$ is monic and a retraction.
2. $\displaystyle f$ is epic and a section.
3. $\displaystyle f$ is an isomorphism.

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 Post subject: Re: Category Theory for beginners (2)Posted: Sat Jan 16, 2016 11:05 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 442
Location: Ioannina, Greece
(I) 1) Because $f:A\longrightarrow B$ is a section, exists $g:B\longrightarrow A$ such that $g\circ f=id_A$.
$$\xymatrix{ A \ar[r]^f \ar[dr]_{id_A} & B \ar[d]^g \\ & A }$$
If for $g_1,g_2:C\longrightarrow A$ holds $f\circ g_1=f\circ g_2$,
$$\begin{xy} (0,0)*+{C}="a", (20,0)*+{A}="b", (40,0)*+{B}="c" \ar @{->}^{g_1} "a";"b" < 2pt> \ar @{->}_{g_2} "a";"b" <-2pt> \ar @{->}^f "b";"c" \end{xy}$$ then \begin{align*}
\end{align*}
So, $f$ is monic.

2) Because $f:A\longrightarrow B$ is a retraction, exists $g:B\longrightarrow A$ such that $f\circ g=id_B$.
$$\xymatrix{ A \ar[r]^{f} & B \\ B \ar[u]^g \ar[ur]_{id_B} & }$$
If for $g_1,g_2:B\longrightarrow C$ holds $g_1\circ f=g_2\circ f$, $$\begin{xy} (0,0)*+{A}="c", (20,0)*+{B}="b", (40,0)*+{ C}="a" \ar @{->}^f "c";"b" \ar @{->}^{g_1} "b";"a" < 2pt> \ar @{->}_{g_2} "b";"a" <-2pt> \end{xy}$$ then \begin{align*}
(g_1\circ f)\circ g =(g_2\circ f)\circ g \quad&\Rightarrow\quad g_1\circ( f\circ g) =g_2\circ( f\circ g)\\
\end{align*}
So, $f$ is epic.

_________________
Grigorios Kostakos

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 Post subject: Re: Category Theory for beginners (2)Posted: Sat Jan 16, 2016 11:54 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
(II) Let $\displaystyle{f\in\rm{Mor}(A,B)}$ where $\displaystyle{A\,,B\in C}$ .

$\displaystyle{i)\implies ii)}$ : Suppose that $\displaystyle{f}$ is monic and retraction.

According to (I) , $\displaystyle{f}$ is epic.

Also, there exists $\displaystyle{g:B\to A}$ such that $\displaystyle{f\circ g=Id_{B}}$.

Now, $\displaystyle{(f\circ g)\circ f=Id_{B}\circ f\implies f\circ (g\circ f)=f=f\circ Id_{A}}$

and since $\displaystyle{f}$ is monic, we get : $\displaystyle{g\circ f=Id_{A}}$, so it is a section.

$\displaystyle{ii)\implies iii)}$ : Since $\displaystyle{f}$ is section, there exists

$\displaystyle{g:B\to A}$ such that $\displaystyle{g\circ f=Id_{A}}$. Now,

$\displaystyle{f\circ (g\circ f)=f\circ Id_{A}\implies (f\circ g)\circ f=f=Id_{B}\circ f}$ and

using the fact that $\displaystyle{f}$ is epic, we get $\displaystyle{f\circ g=Id_{B}}$.

Then, $\displaystyle{f}$ is an isomorphism.

$\displaystyle{iii)\implies i)}$ : There exists $\displaystyle{g:B\to A}$ such that

$\displaystyle{g\circ f=Id_{A}\,\,,f\circ g=Id_{B}}$ .

The first relation gives us that $\displaystyle{f}$ is section, so monic. The second relation

gives us that $\displaystyle{f}$ is retraction

and the exercise comes to an end.

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