# 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

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 AKE, 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