$f$ is surjective iff $g circ f = h circ frightarrow g=h$
$begingroup$
My question is:
Let $f:A rightarrow B$ be a function.
Prove that
$f$ is surjective iff for every pair of functions $g:B rightarrow C $ and $h:B rightarrow C $,
$g circ f = h circ frightarrow g=h$
How can I prove this? Thanks in advance!
functions elementary-set-theory function-and-relation-composition
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
asked Apr 15 '18 at 9:08
DMKDMK
41
$endgroup$
add a comment |
$begingroup$
My question is:
Let $f:A rightarrow B$ be a function.
Prove that
$f$ is surjective iff for every pair of functions $g:B rightarrow C $ and $h:B rightarrow C $,
$g circ f = h circ frightarrow g=h$
How can I prove this? Thanks in advance!
functions elementary-set-theory function-and-relation-composition
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
asked Apr 15 '18 at 9:08
DMKDMK
41
$endgroup$
$begingroup$
Have you got any ideas on how to prove this? You'll get better responses (and fewer down votes) if you show your own progress or thoughts so far.
$endgroup$
– B. Mehta
Apr 15 '18 at 11:34
add a comment |
$begingroup$
My question is:
Let $f:A rightarrow B$ be a function.
Prove that
$f$ is surjective iff for every pair of functions $g:B rightarrow C $ and $h:B rightarrow C $,
$g circ f = h circ frightarrow g=h$
How can I prove this? Thanks in advance!
functions elementary-set-theory function-and-relation-composition
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
asked Apr 15 '18 at 9:08
DMKDMK
41
$endgroup$
My question is:
Let $f:A rightarrow B$ be a function.
Prove that
$f$ is surjective iff for every pair of functions $g:B rightarrow C $ and $h:B rightarrow C $,
$g circ f = h circ frightarrow g=h$
How can I prove this? Thanks in advance!
functions elementary-set-theory function-and-relation-composition
functions elementary-set-theory function-and-relation-composition
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
asked Apr 15 '18 at 9:08
DMKDMK
41
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
asked Apr 15 '18 at 9:08
DMKDMK
41
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
edited Dec 12 '18 at 1:41
Martin Sleziak
44.9k10122277
44.9k10122277
asked Apr 15 '18 at 9:08
DMKDMK
41
asked Apr 15 '18 at 9:08
DMKDMK
41
asked Apr 15 '18 at 9:08
DMKDMK
41
41
$begingroup$
Have you got any ideas on how to prove this? You'll get better responses (and fewer down votes) if you show your own progress or thoughts so far.
$endgroup$
– B. Mehta
Apr 15 '18 at 11:34
add a comment |
$begingroup$
Have you got any ideas on how to prove this? You'll get better responses (and fewer down votes) if you show your own progress or thoughts so far.
$endgroup$
– B. Mehta
Apr 15 '18 at 11:34
$begingroup$
Have you got any ideas on how to prove this? You'll get better responses (and fewer down votes) if you show your own progress or thoughts so far.
$endgroup$
– B. Mehta
Apr 15 '18 at 11:34
$begingroup$
Have you got any ideas on how to prove this? You'll get better responses (and fewer down votes) if you show your own progress or thoughts so far.
$endgroup$
– B. Mehta
Apr 15 '18 at 11:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $f$ is surjective, and $g,h:B to C$ are such that $gneq h$.
We want to prove that $gcirc f neq fcirc g$.
If $gneq f$ then there exists some $b in B$ such that $g(b)neq h(b)$.
As $f$ is surjective, there exists $a in A$ such that $f(a)=b$.
Hence
$$(gcirc f)(a)=g(f(a))=g(b)neq h(b)=h(f(a))=(hcirc g)(a).$$
For the converse, suppose that $f$ is not surjective.
Now we want to find a set $C$ and maps $g,h:Bto C$ such that $gneq h$ but $gcirc f = hcirc f$.
Let $C={0,1}$ and define $g,h:Bto C$ by making
$$
g(x)=0,qquadtext{and}qquad h(x)=
begin{cases}
0 &text{ if } x in f(A)\
1 &text{ otherwise.}
end{cases}
$$
These maps are in the desired conditions.
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Suppose $f$ is surjective, and $g,h:B to C$ are such that $gneq h$.
We want to prove that $gcirc f neq fcirc g$.
If $gneq f$ then there exists some $b in B$ such that $g(b)neq h(b)$.
As $f$ is surjective, there exists $a in A$ such that $f(a)=b$.
Hence
$$(gcirc f)(a)=g(f(a))=g(b)neq h(b)=h(f(a))=(hcirc g)(a).$$
For the converse, suppose that $f$ is not surjective.
Now we want to find a set $C$ and maps $g,h:Bto C$ such that $gneq h$ but $gcirc f = hcirc f$.
Let $C={0,1}$ and define $g,h:Bto C$ by making
$$
g(x)=0,qquadtext{and}qquad h(x)=
begin{cases}
0 &text{ if } x in f(A)\
1 &text{ otherwise.}
end{cases}
$$
These maps are in the desired conditions.
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
$endgroup$
add a comment |
$begingroup$
Suppose $f$ is surjective, and $g,h:B to C$ are such that $gneq h$.
We want to prove that $gcirc f neq fcirc g$.
If $gneq f$ then there exists some $b in B$ such that $g(b)neq h(b)$.
As $f$ is surjective, there exists $a in A$ such that $f(a)=b$.
Hence
$$(gcirc f)(a)=g(f(a))=g(b)neq h(b)=h(f(a))=(hcirc g)(a).$$
For the converse, suppose that $f$ is not surjective.
Now we want to find a set $C$ and maps $g,h:Bto C$ such that $gneq h$ but $gcirc f = hcirc f$.
Let $C={0,1}$ and define $g,h:Bto C$ by making
$$
g(x)=0,qquadtext{and}qquad h(x)=
begin{cases}
0 &text{ if } x in f(A)\
1 &text{ otherwise.}
end{cases}
$$
These maps are in the desired conditions.
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
$endgroup$
add a comment |
$begingroup$
Suppose $f$ is surjective, and $g,h:B to C$ are such that $gneq h$.
We want to prove that $gcirc f neq fcirc g$.
If $gneq f$ then there exists some $b in B$ such that $g(b)neq h(b)$.
As $f$ is surjective, there exists $a in A$ such that $f(a)=b$.
Hence
$$(gcirc f)(a)=g(f(a))=g(b)neq h(b)=h(f(a))=(hcirc g)(a).$$
For the converse, suppose that $f$ is not surjective.
Now we want to find a set $C$ and maps $g,h:Bto C$ such that $gneq h$ but $gcirc f = hcirc f$.
Let $C={0,1}$ and define $g,h:Bto C$ by making
$$
g(x)=0,qquadtext{and}qquad h(x)=
begin{cases}
0 &text{ if } x in f(A)\
1 &text{ otherwise.}
end{cases}
$$
These maps are in the desired conditions.
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
$endgroup$
Suppose $f$ is surjective, and $g,h:B to C$ are such that $gneq h$.
We want to prove that $gcirc f neq fcirc g$.
If $gneq f$ then there exists some $b in B$ such that $g(b)neq h(b)$.
As $f$ is surjective, there exists $a in A$ such that $f(a)=b$.
Hence
$$(gcirc f)(a)=g(f(a))=g(b)neq h(b)=h(f(a))=(hcirc g)(a).$$
For the converse, suppose that $f$ is not surjective.
Now we want to find a set $C$ and maps $g,h:Bto C$ such that $gneq h$ but $gcirc f = hcirc f$.
Let $C={0,1}$ and define $g,h:Bto C$ by making
$$
g(x)=0,qquadtext{and}qquad h(x)=
begin{cases}
0 &text{ if } x in f(A)\
1 &text{ otherwise.}
end{cases}
$$
These maps are in the desired conditions.
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
answered Apr 15 '18 at 14:47
amrsaamrsa
3,8452618
3,8452618
add a comment |
add a comment |
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$begingroup$
Have you got any ideas on how to prove this? You'll get better responses (and fewer down votes) if you show your own progress or thoughts so far.
$endgroup$
– B. Mehta
Apr 15 '18 at 11:34