I'm reading a demonstration stating an function from a non-empty domain which is also injective does have a left inverse, I can see why by drawing graphs of injective functions but not using language. Is there a quick explanation for that?
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asked Feb 5, 2015 at 11:59
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2 Answers
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Yes. Let $f\,:\,X\longrightarrow Y$ be injective. Fix $x_0\in X$. For $y\in Y$, define $$g(y)=\begin{cases} x&\text{if}\ y=f(x),\\ x_0&\text{if}\ y\neq f(x)\ \forall x\in X. \end{cases}$$ This can be seen to be a well-defined function $Y\longrightarrow X$ since $f$ is injective, and by construction is a left inverse.
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$\begingroup$ Indeed I see how it works for $x\rightarrow x^2$, $\mathbb{R}^+ \rightarrow \mathbb{R}$ with inverse $\sqrt{x}$ for $x \in \mathbb{R}^+$ otherwise any positive constant (implying that the inverse is not injective). $\endgroup$ Feb 5, 2015 at 12:07
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1$\begingroup$ @Learningisamess: Consider the map $f\colon\{1\}\to\{0,1\}$ defined by $f(1)=1$. Can it have an injective inverse? $\endgroup$– Asaf Karagila ♦Feb 5, 2015 at 12:30
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1$\begingroup$ @Learningisamess: So the inverse is not necessarily injective. It's a much much simpler example. $\endgroup$– Asaf Karagila ♦Feb 5, 2015 at 12:33
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2$\begingroup$ @Learningisamess Actually the inverse is surjective. From equality $s\circ i=\operatorname{id}$ (put this expression somewhere in your memory) you are allowed to conclude that $s$ is surjective and $i$ is injective. Moreover if $s$ is surjective then it has a right inverse, and if $i$ is injective and has a non-empty domain then it has a left inverse. Functions having an empty domain are (vacuously) injective but (alas) have not left inverse unless their codomain is also empty. $\endgroup$– drhabFeb 5, 2015 at 12:51
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Let $f$ be a function from $\varnothing$ to a nonempty set $B$. Then, $f$ is injective, as the statement $(\forall x,y\in\varnothing)(f(x)=f(y)\implies x=y)$ is vacuously true. However, $f$ does not have a left-inverse, as it is not possible for there to be a function from $B$ to $\varnothing$ (where would the elements of $B$ map to?).
However, excluding this trivial case, the statement is true, and the proof can be found in Jason's answer. It is instructive to consider where his argument breaks down in the case of functions with empty domains.
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$\begingroup$ Why do we need to classify these two cases? If $f$ be a function from $\emptyset$ to some other set, the only choice for $f$ is $\emptyset$, so $f$ has left inverse is vacuously true? $\endgroup$ Oct 15, 2023 at 22:55
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$\begingroup$ @Andrew: It's a little difficult to answer your question in a comment. Can I ask what definition of "function" you are using? $\endgroup$– JoeOct 16, 2023 at 11:19
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$\begingroup$ See my question math.stackexchange.com/q/4787600/1055938 $\endgroup$ Oct 16, 2023 at 12:43