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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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    I've added diagrams which I created in LaTeX using [xypic](http://en.wikibooks.org/wiki/LaTeX/Creating_Graphics#Xy-pic) package and then I've converted them using dvipng. I guess we do not have here some easier option to make diagrams than including pictures.2011-11-26
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    @Martin: I hope you do not mind, but I removed the diagrams in favor of these (rather ugly, ad-hoc) ones because the image hosting site eventually removes the images, and we want the questions to be understandable even after that occurs.2011-11-26
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    @Zev: Of course I don't mind, do anything what you consider an improvement. Based on [this comment](http://meta.math.stackexchange.com/questions/3130/using-other-peoples-pictures/3178#comment12061_3178) I thought that it is ok to use images. The comment says: *Images are now uploaded to a Stack Exchange server (stack.imgur.com) where images do not expire.*\\ Anyway, if there is a danger that images will die eventually, perhaps for now both form of diagrams could coexist in the question...?2011-11-26
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    I do not believe it is the case that images on stack.imgur.com don't expire, my impression was that they are only kept up for a longer time (also, I'm afraid I don't see the comment you're referring to?). As the comments on Jeff's answer there indicate, they do seem to expire after a few months.2011-11-26
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    That is certainly fair, I have included both.2011-11-26
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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