# Introduction

In Basic Algebra I, I am struggling with fully understanding the following exercise:

Show that $S\overset{\alpha}{\to}T$ is injective if and only if there is a map $T\overset{\beta}{\to}S$ such that $\beta\alpha=1_S$, surjective if and only if there is a map $T\overset{\beta}{\to}S$ such that $\alpha\beta=1_T$. In both cases, investigate the assertion: if $\beta$ is unique then $\alpha$ is bijective.

# My Problem

I am struggling only with the bold portion. (I have written proofs by contradiction for the other aspects of the question.) What confuses me specifically is this:

• What is this question really asking? Is it saying, "What happens when $\beta$ is unique when both $\beta\alpha=1_S$ and $\alpha\beta=1_T$?" or is it saying, "What happens when $\beta$ is unique and either $\beta\alpha=1_S$ or $\alpha\beta=1_T$ is true?"

# Remarks

As you can see, my real problem here is understanding precisely what is being asked. If it is asking, the first (both $\alpha\beta=1_T$ and $\beta\alpha=1_S$ are true), then we're simply constructing the very definition of a bijection. If it's asking the latter, I don't know what's going on . . . Are we somehow still constructing a bijection?

Can you all give me help on reading questions such as this?

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That an injective map has a left inverse is only true if its domain is nonempty. –  Michael Greinecker Dec 29 '12 at 21:01
@MichaelGreinecker I agree mainly, but there is one exception. An injective map also has a left-inverse if its domain and codomain are both empty. –  drhab Apr 23 at 18:51
@drhab Point taken. –  Michael Greinecker Apr 24 at 5:00
See the posts linked to this one: math.stackexchange.com/questions/553186/… –  Martin Sleziak Jul 3 at 18:08

I parse this as follows:

The bold sentence refers separately to each of the two statements.

So expanded out, this would be:

1a. Show that ... is injective if and only if ... Show also that if $\beta$ is unique then $\alpha$ is bijective.

1b. Show that ... is surjective if and only if... Show also that if $\beta$ is unique then $\alpha$ is bijective.

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Why do you parse this this way? Why is this correct when the other interpretation is not? I'm not saying you're wrong. I'm genuinely asking those questions because I don't see why this is the intended meaning of the question. –  000 Dec 29 '12 at 1:00
Fair question. I replace 'both' with 'each', where the claim is that the statement "in both cases" is true iff "each case" is true separately. –  Assad Ebrahim Dec 29 '12 at 1:04
Could you explain what the mathematical consequences of your interpretation are? I am not seeing any and it makes the question seem null. I am most likely wrong as I am heavily confused. I think it is interpreted the other way because that appears to be the only way of making a sensible question. I mean, Jacobson (the author) even puts a definition of bijections in terms identical to what the other interpretation says: "$S\overset{\alpha}{\to}T$ is bijective if and only if there exists a map $T\overset{\beta}{\to}S$ such that $\beta\alpha=1_S$ and $\alpha\beta=1_T$." –  000 Dec 29 '12 at 1:09
Jacobson (in my opinion) is not the clearest author, and I've also found understanding exactly what he means a tad challenging (frustrating?) Do you want to take this into a chat room? It may be quicker and easier there. –  Assad Ebrahim Dec 29 '12 at 1:17
Transcript resolving the problem. –  Assad Ebrahim Dec 29 '12 at 2:15

# Central Matter

With the help of Assad Ebrahim, I was able to figure out what's being said here. The crux is this: $$\beta \text{ is unique} \iff |S|=|T| \iff \alpha \text{ is bijective},$$ which quickly implies that if $\alpha$ is bijective in either case.

# Elaboration and Specifics

For the mapping $\alpha$ wherein $\beta\alpha=1_S$, we have that $\alpha$ is injective. However, $\beta$ can be any map $T\to S$ such that all the elements of $T$ map to $S$. This means that $\beta$ acts not just on the elements $\alpha(s)$; it acts on all elements of $T$. As a result, there are, in general, elements in $T$ which can be mapped to any $s\in S$. Thus, there are many possible maps $\beta$; we can create a new one simply by changing what a given $t\in T$ which is not $\alpha(s)$ maps to.

Now, the issue is this: We are supposing $\beta$ is unique. This means there cannot be elements in $T$ which are not equal to $\alpha(s)$. If there were, then $\beta$ would cease to be unique for the reason outlined just above. Hence, we have that $\alpha$ is also surjective. Therefore, $\alpha$ is bijective. $\blacksquare$

For the mapping $\alpha$ wherein $\alpha\beta=1_T$, we have a similar situation. Using the same line of logic, we see that there cannot be $s\in S$ such that $s\ne \beta(t)$. If there were, $\beta$ would cease to be unique. Thus, we have that $\alpha$ is injective. Therefore, $\alpha$ is bijective. $\blacksquare$

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Hint: You might want to start with extreme cases, such as when $S$ or $T$, but not both, has only one element. (I take the problems to be whether "$\alpha$ is bijective if and only if there is a unique map $\beta$ such that $\beta\alpha=1_S$" and similarly for the other part.) –  Michael E2 Dec 29 '12 at 5:23
@MichaelE2 That's a very good point. This question has indeed taught me to examine cases in particular detail. –  000 Dec 29 '12 at 5:49
If $S=\{s\}$ (a singleton) then there is exactly one map $\beta:T\rightarrow S$ (so it is unique) and $\beta\alpha=1_S$. If $T$ contains other elements next to $\alpha(s)$ (wich is the case if $T$ is not a singleton) then $\alpha$ is not surjective. So uniqueness of $\beta$ does not imply that $\alpha$ is bijective. –  drhab Apr 23 at 18:35

If $\beta:T\rightarrow S$ exists with $\beta\alpha=id_{S}$ then we find easily that $\alpha\left(s\right)=\alpha\left(s'\right)$ implies that $s=\beta\alpha\left(s\right)=\beta\alpha\left(s'\right)=s'$ showing that $\alpha$ is injective.

Conversely let $\alpha:S\rightarrow T$ be injective. If $s_{0}\in S$ then we can construct $\beta:T\rightarrow S$ by sending each element $t=\alpha\left(s\right)\in\alpha\left(S\right)$ to the unique $t\in T$. Elements not contained in $\alpha\left(S\right)$ can be sent to $s_{0}$. Then $\beta\alpha=id_{S}$. Note however that this does not have to work if $S=\emptyset$. We have the unique empty map $\alpha:\emptyset\rightarrow T$ wich is vacuously injective. Only if also $T=\emptyset$ then there exist the empty map $\beta:T=\emptyset\rightarrow\emptyset$ and indeed $\beta\alpha=id_{\emptyset}$. However, if $T\ne\emptyset$ then no map $\beta:T\rightarrow\emptyset$ exists.

So the statement: if $\alpha:S\rightarrow T$ is injective then $\beta\alpha=id_{S}$ for some $\beta:T\rightarrow S$, is true under the extra condition that $S\ne\emptyset\vee T=\emptyset$.

If $S$ is a singleton then automatically $\beta$ is unique. However, if moreover $T$ is not a singleton then $\alpha$ is not a bijection. So uniqueness of $\beta$ does not imply that $\alpha$ is bijective. For this we need the extra condition that $S$ is no singleton or $T$ is a singleton.

If $\alpha\beta=id_{T}$ then $t=\alpha\left(\beta\left(t\right)\right)$ for each $t\in T$ showing immediately that $\alpha$ is surjective.

Conversely let $\alpha:S\rightarrow T$ be surjective. Then we construct $\beta:T\rightarrow S$ by sending each element $t$ by one of an elements $s\in S$ that suffices $\alpha\left(s\right)=t$. Then automatically $\alpha\beta=id_{T}$. Note that this equation implies that $\beta(t)$ belongs to fibre $\alpha^{-1}\left(\left\{ t\right\} \right)$, so that is necessary.

If this $\beta$ with $\alpha\beta=id_{T}$ is unique then for each $t\in T$ the fibre $\alpha^{-1}\left(\left\{ t\right\} \right)$ contains exactly one element. This tells us that the surjective $\alpha$ is also injective, hence bijective.

In category Sets every epimorphism (surjection) is a retraction (second case), but not every monomorphism (injection) is a section (first case). Exceptions are the elements in $\mathbf{Sets}\left(\emptyset,T\right)$ where $T\ne\emptyset$. They are injective but are not sections.

I have been working with sets. If you are working in groups, rings, et cetera then the underlying sets will not be empty.

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