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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 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. – Martin Sleziak Nov 26 '11 at 8:49
@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. – Zev Chonoles Nov 26 '11 at 8:53
@Zev: Of course I don't mind, do anything what you consider an improvement. Based on this comment I thought that it is ok to use images. The comment says: *Images are now uploaded to a Stack Exchange server ( 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...? – Martin Sleziak Nov 26 '11 at 8:59
I do not believe it is the case that images on 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. – Zev Chonoles Nov 26 '11 at 9:03
That is certainly fair, I have included both. – Zev Chonoles Nov 26 '11 at 9:09
up vote 2 down vote accepted

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? – Kodak Nov 27 '11 at 6:59

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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