# Not quite isomorphisms?

Sorry, I'm back with the elementary questions. I've been reading up on the introduction to category theory, and paused over the definition of an isomorphism:

An isomorphism in a category is a morphism $f \colon X \rightarrow Y$ for which there exists a morphism $g \colon Y \rightarrow X$ so that $g f = 1_X$ and $f g = 1_Y$.

However, knowing the congenital sneakiness of mathematicians, presumably there are cases where only 'one side' is true, eg. $g f = 1_X \wedge f g \neq 1_Y$? Or maybe put better, $g f = 1_X$ does not imply $f g = 1_Y$?

• "congenital sneakiness of mathematicians" made my day. May 30, 2018 at 10:37
• You can easily answer your question in the category of sets. What does $gf = 1_X$ mean? May 30, 2018 at 10:41

Yes. Suppose that we are working in the category of sets and that the morphisms are the functions. Consider a set $X$ with a single element $a$ and a set $Y$ with more than one element, one of which is $a$. Define $f\colon X\longrightarrow Y$ by $f(a)=a$ and let $g\colon Y\longrightarrow X$ be the constant function $a$. Then $g\circ f=1_X$, but $f\circ g\neq1_Y$.

You are absolutely right, such pairs $f, g$ do indeed exist.

As a particular example, take $f, g : \mathbb N \rightarrow \mathbb N$ (in the category Set of sets and functions), defined by

$$f(n) = n+1$$

$$g(n) = \begin{cases} 0 \,\, \text{if} \,\,n=0 \\ n-1 \,\, \text{otherwise} \end{cases}$$

Then,

$$(g \circ f)(n) = g(f(n)) = n = \operatorname{id}_{\mathbb N}(n)$$ but

$$(f\circ g)(n) = \begin{cases} 1 \,\, \text{if} \,\,n=0 \\ n \,\, \text{otherwise} \end{cases}$$

which is clearly ${\color{red}\neq}\operatorname{id}_{\mathbb N}(n)$

If $f:X\to Y$ and $g:Y\to X$ are morphisms such that $f\circ g=\text{id}_Y$, then we call $g$ a section of $f$ and $f$ a retraction of $g$.

A standard case is sections of a vector bundle $(E,B,\pi)$ where $\pi:E\to B$ is the projection map, and a section is a map $s:B\to E$ such that $\pi\circ s=\text{id}_B$.

• No, $f$ is a retraction. In this case, $Y$ is a retract of $X$.
– Pece
May 30, 2018 at 10:59
• For a beginner, as OP claims to be, your reply is not very helpful. A beginner would benefit from a more accessible example; rather than one involving advanced abstract algebra. Jun 2, 2018 at 2:59
• @MusaAl-hassy: People come from lots of different backgrounds. Incidentally, I would have classified the example as more along the lines of "basic differential geometry" (maybe even basic topology) rather than "advanced abstract algebra".
– user14972
Jun 3, 2018 at 2:33

A more illuminating definition of $$X\xrightarrow{f}Y$$ being an isomorphism is that it has both a right inverse and a left inverse.

Definition. Given a morphism $$X\xrightarrow{f}Y$$, we say a morphism $$X\xleftarrow{f^{-1}}Y$$ is a right inverse (also called a section) of $$X\xrightarrow{f}Y$$ if $$f\circ f^{-1}=\mathrm{id}_Y$$. Similarly, we say $$X\xleftarrow{{}^{-1}}Y$$ is a left inverse (also called a retraction) of $$X\xrightarrow{f}Y$$ if $${}^{-1}f\circ f=\mathrm{id}_X$$. A morphism $$X\xrightarrow{f}Y$$ is an isomorphism if it has both a left inverse and a right inverse.

Example 1. Most functions labeled inverses in basic mathematics are actually right inverses: $$\mathbb R_{\geq0}\xrightarrow{\sqrt(x)}\mathbb R$$ is a right inverse of $$\mathbb R\xrightarrow{x^2}\mathbb R_{\geq0}$$ because $$(\sqrt{x})^2=x$$, $$(-1,1)\xrightarrow{\arcsin(x)}\mathbb R$$ is a right inverse of $$\mathbb R\xrightarrow{\sin(x)}(-1,1)$$ because $$\sin(\arcsin(x))=x$$. These right inverses are also not unique: we also have $$(-\sqrt{x})^2=x$$, $$\sin(\pi-\arcsin(x))=x$$, and $$\sin(\arcsin(x)+2\pi)=x$$.

Lemma. If $$X\xrightarrow{f}Y$$ has both a left and a right inverse (i.e. if it is an isomorphism), then the two inverses coincide, i.e. $$f^{-1}={}^{-1}f$$. In particular, if $$X\xrightarrow{f}Y$$ is an isomorphism, it has a unique left inverse and a unique right inverse, these coincide, and are themselves an isomorphism.

Example 2. Since $$\mathbb R\xrightarrow{x^2}\mathbb R_{\geq0}$$ has multiple distinct right inverses, it is not an isomorphism. Consequently, its right inverse $$\mathbb R\xleftarrow{\sqrt x}\mathbb R_{\geq0}$$ is also not an isomorphism.

Remark. For general categories it is not enough for a morphism $$X\xrightarrow{f}Y$$ to have a unique right or left inverse for it to be an isomorphism, but in some categories (like the category of sets, and I think the category of vector spaces) this suffices.