Direct Products of Modules .... Canonical Injections ....

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with some aspects of the proof of Proposition 2.1.4 ...

Proposition 2.1.4 and its proof read as follows:

In the above proof by Paul Bland we read the following:

" ... ... Since $p_\alpha u_\alpha = \text{ id}_{ M_\alpha }$, we have that $u_\alpha$ is an injective mapping and that $p_\alpha$ is surjective ... ... "Can someone please explain exactly how/why $u_\alpha$ is an injective mapping ... ?Help will be appreciated ...

Peter

 

[member 587159]

A mapping is injective if and only if there exists a left inverse for the map. Here, the left inverse is given by $p_\alpha$.

Also, a mapping is surjective if there exists a right inverse for the map.

As a side note, the converse implication for this last statement also holds, if and only if the axiom of choice is true, but you won't need it here.

Can you prove these 2 claims?

 

[Math Amateur]

Hi Math_QED ...

Note the following:

Given $f \ : \ A \longrightarrow B$, we say that a function $g \ : \ B \longrightarrow A$ is a left inverse for $f$ if $g \circ f = i_A$

Will try to prove the following:

$f$ has a left inverse $\Longleftrightarrow f$ is injective Now assume $f$ has a left inverse, say, $g$ ...

Let $f(a) = f(b)$ ... ... need to show $a = b$ ...

Now $g \circ f (a) = g( f(a) ) = i_A (a) = a$ ... ... ... ... ... (1)

and $g \circ f (b) = g( f(b) ) = i_A (b) = b$ ... ... ... ... ... (2)But since $f(a) = f(b)$ ...

... we have that (2) $\Longrightarrow g \circ f (b) = b = g( f(b) ) = g( f(a) ) = i_A (a) = a$

... so $a =b$ ... ...

Is that correct ...?

NOTE ... now need to show that $f$ is injective $\Longrightarrow f$ has a left inverse ...

BUT ... not making any meaningful progress ... can you help ...?

Peter

 

[member 587159]

Hi Math_QED ...

Note the following:

Given $f \ : \ A \longrightarrow B$, we say that a function $g \ : \ B \longrightarrow A$ is a left inverse for $f$ if $g \circ f = i_A$

Will try to prove the following:

$f$ has a left inverse $\Longleftrightarrow f$ is injectiveNow assume $f$ has a left inverse, say, $g$ ...

Let $f(a) = f(b)$ ... ... need to show $a = b$ ...

Now $g \circ f (a) = g( f(a) ) = i_A (a) = a$ ... ... ... ... ... (1)

and $g \circ f (b) = g( f(b) ) = i_A (b) = b$ ... ... ... ... ... (2)But since $f(a) = f(b)$ ...

... we have that (2) $\Longrightarrow g \circ f (b) = b = g( f(b) ) = g( f(a) ) = i_A (a) = a$

... so $a =b$ ... ...

Is that correct ...?

NOTE ... now need to show that $f$ is injective $\Longrightarrow f$ has a left inverse ...

BUT ... not making any meaningful progress ... can you help ...?

Peter

Your first proof is correct. Well done!

The other direction is not necessary in the proof you are working on, and it is the harder one. But it is an interesting and basic theorem anyway.

We have to actually construct a left inverse for $f$, and once you have seen the trick, you will be able to apply it to similar problems.

So, our goal is to make a function $g: B \to A$ that 'undoes' $f$ from the left.

How can we map an element $b \in B$ to an element in $a \in A$ in a manner that has something to do with $f$?

Hint: $f(A) \subseteq B$

 

[Math Amateur]

Your first proof is correct. Well done!

That direction is not necessary in the proof, and it is the harder one. But it is an interesting and basic theorem anyway.

We have to actually construct a left inverse for $f$, and once you have seen the trick, you will be able to apply it to similar problems.

So, our goal is to make a function $g: B \to A$ that 'undoes' $f$ from the left.

How can we map an element $b \in B$ to an element in $a \in A$ in a manner that has something to do with $f$?

Hint: $f(A) \subseteq B$

Given $f \ : \ A \longrightarrow B$ is injective ... we want to show that $f$ has a left inverse function, say $g$ ...

Define $g$ this way ... where $f(a) = b$ for $a \in A, b \in B$ ... define $g$ such that $g \ : \ B \longrightarrow A$ where $g(b) = a$ ...

... so that $g \circ f (a) = g(f(a) ) = a$ ...

The above definition is possible because $f$ is injective ...

Defining $g$ in the above way for each $a \in A, b \in B$ where $f(a) = b$ means $g$ is the left inverse of $f$ ...Is that correct?

Peter

 

[member 587159]

Given $f \ : \ A \longrightarrow B$ is injective ... we want to show that $f$ has a left inverse function, say $g$ ...

Define $g$ this way ... where $f(a) = b$ for $a \in A, b \in B$ ... define $g$ such that $g \ : \ B \longrightarrow A$ where $g(b) = a$ ...

... so that $g \circ f (a) = g(f(a) ) = a$ ...

The above definition is possible because $f$ is injective ...

Defining $g$ in the above way for each $a \in A, b \in B$ where $f(a) = b$ means $g$ is the left inverse of $f$ ...Is that correct?

Peter

The idea is correct, but your function is not defined on $B$ but rather on $f(A)$. What if $f$ is not surjective? Then some $b$'s won't be attained by $f$.

Just this needs fixing. Can you think of a way to give an image to $b \notin f(A)$?

You noted correctly that injectivity is required to make the function $g$ well-defined.

 

[Math Amateur]

Hmm ... can see the problem ...

But cannot see the answer to the problem ...

Peter

 

[member 587159]

Hmm ... can see the problem ...

But cannot see the answer to the problem ...

Peter

In the end, you want that for $a \in A$,

$g \circ f(a) = g(f(a)) = a$

So, for this condition to be true, it doesn't matter how $g$ is defined outside $f(A)$.

Can you have another attempt now?

 

[Math Amateur]

Well ... if how $g$ is defined outside $f(A)$ just put $g(b) = 0$ or simply $g(b)$ = arbitrary element of $A$ if $0$ is not an element of $A$ ... and do this for all $b$ outside $f(A)$ ...

Would that do ...?

Peter

 

[member 587159]

Well ... if how $g$ is defined outside $f(A)$ just put $g(b) = 0$ or simply $g(b)$ = arbitrary element of $A$ if $0$ is not an element of $A$ ... and do this for all $b$ outside $f(A)$ ...

Would that do ...?

Peter

Well, an arbitrary fixed element indeed works!

 

[Math Amateur]

Thanks Math_QED ... really appreciate your help...

Peter

 

[mathwonk]

I agree with everything here. But I would like to remark that the situation here, as QED well knows, is a little more subtle. In this proof the left inverse that is given is also linear. In fact an injective linear map, although it has a left inverse function, need not have one that is linear. So since this injective linear map has a linear left inverse, it is a little more special than just any old injective linear map. E.g. the injection of the even integers into the group of all integers, is injective and linear, but has no linear left inverse. Just didn't want the OP to come away with a possible wrong impression, but maybe no need to worry.

 

[Math Amateur]

Thanks mathwonk ... appreciate your help, guidance and support...

Peter