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 August 7th, 2014, 05:23 PM #1 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 Show that there exists a unique mapping $g:B \rightarrow A$, such that $gof=i_A$ and please,check my answer to the folowing problem: "Let $i_A:A\rightarrow A$ and $i_B:B \rightarrow B$ be two identity mappings and let $f:A \rightarrow B$ be a bijection. Show that there exists a unique mapping $g:B \rightarrow A$, such that $gof=i_A$ and $fog=i_B$". my solution: (1) $f$ is onto, then for every $y \in B$ there exists at least one $x \in A$ such that $f(x)=y$.For this "x", consider there exists also $g':B \rightarrow A$ such that $g'of=i_A$. Then $g'(f(x))=g(f(x))=x \rightarrow g'=g$. (2) $f$ is injective. Then $f(x_1)=f(x_2) \rightarrow x_1=x_2$. Consider $g':B \rightarrow A$ such that $fog'=i_B$. Then $f(g'(y))=f(g(y))=y \rightarrow g'=g$. Thanks! August 8th, 2014, 01:54 AM   #2
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 Originally Posted by walter r please,check my answer to the folowing problem: "Let $i_A:A\rightarrow A$ and $i_B:B \rightarrow B$ be two identity mappings and let $f:A \rightarrow B$ be a bijection. Show that there exists a unique mapping $g:B \rightarrow A$, such that $gof=i_A$ and $fog=i_B$". my solution: (1) $f$ is onto, then for every $y \in B$ there exists at least one $x \in A$ such that $f(x)=y$.For this "x", consider there exists also $g':B \rightarrow A$ such that $g'of=i_A$. Then $g'(f(x))=g(f(x))=x \rightarrow g'=g$. (2) $f$ is injective. Then $f(x_1)=f(x_2) \rightarrow x_1=x_2$. Consider $g':B \rightarrow A$ such that $fog'=i_B$. Then $f(g'(y))=f(g(y))=y \rightarrow g'=g$. Thanks!
This seems backwards to me.

For (1), since for every $y \in B$ we have $x \in A$ with $f(x) = y$, we want to DEFINE:

$g(y) = x$.

Note that $g$ is a choice function for the collection of non-empty sets:

$\displaystyle \bigcup_{y \in B} f^{-1}(y)$

so an explicit definition of $g$ requires the Axiom of Choice.

Then $(f\circ g)(y) = f(x) = y$, for all $y \in B$, that is: $f \circ g = 1_B$.

The $g$ so obtained is by no means necessarily unique.

For (2) the situation is much nicer: we can define $g: f(A) \to A$ by:

$g(f(x)) = x$. This is well-defined, since if $f(x_1) = f(x_2)$, we have $x_1 = x_2$, so that $g$ so defined is indeed a function.

We have, therefore $g \circ f = 1_A$. This $g$ is uniquely defined, but only on $f(A)$.

If $f$ is onto, then $f(A) = B$, so we can use the $g$ of (2) (thereby circumventing the need for the Axiom of Choice). For this $g$, we clearly have:

$g \circ f = 1_A$.

Now, for any $y \in B$ we have a unique $x \in A$ with $f(x) = y$ (since $f$ is injective and surjective...the existence of $x$ follows because $f$ is surjective, uniqueness follows since $f$ is injective), hence:

$(f \circ g)(y) = (f \circ g)(f(x)) = ((f \circ g )\circ f)(x) = (f \circ (g \circ f))(x) = (f \circ 1_A)(x) = f(x) = y$

so that $f \circ g = 1_B$,

This shows we have at least one such function $g$. Suppose we have another, $g'$.

Then $g' = g' \circ 1_B = g' \circ (f \circ g) = (g' \circ f) \circ g = 1_A \circ g = g$, so $g$ is unique.

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The problem with your answer is that you assume the existence of the $g$ you are trying to prove exists, and you never show your two $g$'s are the same function.

Furthermore, it IS NOT TRUE that for $f$ onto, there IS a $g$ such that $g \circ f = 1_A$.

To see this, suppose that $A = \{a,b\}$, and that $B = \{c\}$, and that:

$f(x) = c$, for all $x \in A$ ($f$ is a constant function).

There are only two possible functions $g: B \to A$, namely:

$g_1(c) = a$, and
$g_2(c) = b$.

Now $(g_1 \circ f)(b) = g_1(c) = a \neq b$, while $(g_2 \circ f)(a) = g_2(c) = b \neq a$, so there is NO function $g: B \to A$ with $g\circ f = 1_A$.

Furthermore, if $f$ is injective, there may not exist a function $g: B \to A$ with $f \circ g = 1_B$. For example, let $A = \{a\}$ and let $B = \{u,v\}$.

Define $f(a) = u$. This is injective.

The only possible function $g: B \to A$ is $g(y) = a$, for all $y \in B$.

Now $(f \circ g)(v) = f(g(v)) = f(a) = u \neq v$.

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Surjective means a RIGHT inverse exists (every co-domain element has a pre-image that can be re-mapped to its image).

Injective means a LEFT inverse exists (we can "undo $f$"). August 8th, 2014, 04:30 PM #3 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 excelent explanation. Thanks a lot! Tags $gb,$gofia\$, exists, mapping, rightarrow, show, unique Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Sebastian Garth Number Theory 5 April 29th, 2014 09:01 AM annakar Linear Algebra 0 January 10th, 2013 09:58 AM 450081592 Real Analysis 1 October 16th, 2011 12:26 AM mysterious_poet_3000 Real Analysis 1 February 8th, 2011 03:17 PM malaizai Algebra 23 April 2nd, 2010 01:38 AM

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