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August 7th, 2014, 06:23 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, 02:54 AM  #2  
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Quote:
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 nonempty 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 welldefined, 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. ********************* 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 $(f \circ g)(v) = f(g(v)) = f(a) = u \neq v$. ******************* Surjective means a RIGHT inverse exists (every codomain element has a preimage that can be remapped to its image). Injective means a LEFT inverse exists (we can "undo $f$").  
August 8th, 2014, 05:30 PM  #3 
Member Joined: Mar 2013 Posts: 71 Thanks: 4 
excelent explanation. Thanks a lot!


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$gb, $gofia$, exists, mapping, rightarrow, show, unique 
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