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The textbook I'm using says that a morphism might have only a section, or only a retraction, but I can't work out an example.

Take objects $A$ and $B$, and morphisms $f$ and $g$,

nice diagram: $\hskip1in$ diag1

ugly diagram: $\matrix{B \hspace{-0.11in} & & \hspace{-0.1in}\xrightarrow{\quad\text{Id}_B\quad} & &\hspace{-0.1in} B\\ & g\searrow& \hspace{-1in}& \nearrow f\\ & \text{}\hskip{-1in}& A &}$

So according to my book, $g$ is a section for $f$ because $g;f =\text{Id}_B$

But that can be rearranged to this just be removing the $\text{Id}_B$ and adding an $\text{Id}_A$:

nice diagram: $\hskip1in$ diag2

ugly diagram: $\matrix{A \hspace{-0.11in} & & \hspace{-0.1in}\xrightarrow{\quad\text{Id}_A\quad} & &\hspace{-0.1in} A\\ & f\searrow& \hspace{-1in}& \nearrow g\\ & \text{}\hskip{-1in}& B &}$

So $g$ is a retraction for $f$ because $f;g = \text{Id}_A$.

It seems to me that this would apply everywhere, and so any morphism that has a section must also have a retraction. What am I missing?

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    So, is it the case that even though id_B is its own inverse, we can't add that inverse to the diagram? So we have to work with algebra and merely communicate with diagrams? Have I understood the situation?2017-08-30

2 Answers 2

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It can't be rearranged - you can write down that second triangle, sure, but there is no guarantee that it is a commutative triangle.

A simple example is if $A=\{a,b\}$ and $B=\{c\}$, then we could define $f:A\to B$ and $g:B\to A$ by $f(a)=c,\quad f(b)=c,\quad g(c)=a$ and they will satisfy $f\circ g=\text{Id}_B$, but $(g\circ f)(b)=g(c)=a\neq b$, so $g\circ f\neq\text{Id}_A$.

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    This helped me realize that I was interpreting the diagram too broadly. The (first) diagram represents $g;f =\text{Id}_B$ so there is an $=$ implied in the diagram. When I redrew the diagram by collapsing and expanding identities, I was moving the implied $=$ as well, which I can't do. Is that accurate? Is there a convention for conveying the location of the $=$ in such a diagram?2011-11-27
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It is not true that $g\circ f=id_B$ implies $f\circ g=id_A$, you can find easy counterexamples already in the category Set.

In Set a map is injective if and only if it has left inverse - see proofwiki. A map is surjective if and only if it has right inverse, with some exceptions concerning empty set - see proofwiki.

I believe you can find easily example of function that is injective and not surjective and vice-versa.

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    I don't the the qeustioner was asking "is my conclusion correct" but rather, why has he been able to conclude this from the diagram when he think's he shouldn't have been able to.2017-08-30