Mixing
What can be said about a function in this case?
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Suppose I have self-maps $g$ and $h$ from set $X$ to $X$ such that $g$ $circ$ $h$ is a constant function and $h$ is surjective. What can be said about function $g$?
I think that g should also be a constant function, but I do not know how to prove that? Any help or tips?
abstract-algebra functions proof-writing
asked Feb 3 at 2:26
UfomammutUfomammut
393314
$endgroup$
add a comment |
$begingroup$
Suppose I have self-maps $g$ and $h$ from set $X$ to $X$ such that $g$ $circ$ $h$ is a constant function and $h$ is surjective. What can be said about function $g$?
I think that g should also be a constant function, but I do not know how to prove that? Any help or tips?
abstract-algebra functions proof-writing
asked Feb 3 at 2:26
UfomammutUfomammut
393314
$endgroup$
add a comment |
$begingroup$
Suppose I have self-maps $g$ and $h$ from set $X$ to $X$ such that $g$ $circ$ $h$ is a constant function and $h$ is surjective. What can be said about function $g$?
I think that g should also be a constant function, but I do not know how to prove that? Any help or tips?
abstract-algebra functions proof-writing
asked Feb 3 at 2:26
UfomammutUfomammut
393314
$endgroup$
Suppose I have self-maps $g$ and $h$ from set $X$ to $X$ such that $g$ $circ$ $h$ is a constant function and $h$ is surjective. What can be said about function $g$?
I think that g should also be a constant function, but I do not know how to prove that? Any help or tips?
abstract-algebra functions proof-writing
abstract-algebra functions proof-writing
asked Feb 3 at 2:26
UfomammutUfomammut
393314
asked Feb 3 at 2:26
UfomammutUfomammut
393314
asked Feb 3 at 2:26
UfomammutUfomammut
393314
asked Feb 3 at 2:26
UfomammutUfomammut
393314
asked Feb 3 at 2:26
UfomammutUfomammut
393314
393314
add a comment |
add a comment |
1 Answer
1
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Yes, $g$ is constant. For any $xin X$, there exists $yin X$ with $h(y)=x$, since $h$ is surjective. But we know that $g(x)=g(h(y))=c$ where $c$ is the appropriate constant. Since this is true for all $x$, we have that $g$ is constant.
answered Feb 3 at 2:39
jgonjgon
16.6k32244
$endgroup$
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, $g$ is constant. For any $xin X$, there exists $yin X$ with $h(y)=x$, since $h$ is surjective. But we know that $g(x)=g(h(y))=c$ where $c$ is the appropriate constant. Since this is true for all $x$, we have that $g$ is constant.
answered Feb 3 at 2:39
jgonjgon
16.6k32244
$endgroup$
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
add a comment |
$begingroup$
Yes, $g$ is constant. For any $xin X$, there exists $yin X$ with $h(y)=x$, since $h$ is surjective. But we know that $g(x)=g(h(y))=c$ where $c$ is the appropriate constant. Since this is true for all $x$, we have that $g$ is constant.
answered Feb 3 at 2:39
jgonjgon
16.6k32244
$endgroup$
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
add a comment |
$begingroup$
Yes, $g$ is constant. For any $xin X$, there exists $yin X$ with $h(y)=x$, since $h$ is surjective. But we know that $g(x)=g(h(y))=c$ where $c$ is the appropriate constant. Since this is true for all $x$, we have that $g$ is constant.
answered Feb 3 at 2:39
jgonjgon
16.6k32244
$endgroup$
Yes, $g$ is constant. For any $xin X$, there exists $yin X$ with $h(y)=x$, since $h$ is surjective. But we know that $g(x)=g(h(y))=c$ where $c$ is the appropriate constant. Since this is true for all $x$, we have that $g$ is constant.
answered Feb 3 at 2:39
jgonjgon
16.6k32244
answered Feb 3 at 2:39
jgonjgon
16.6k32244
answered Feb 3 at 2:39
jgonjgon
16.6k32244
answered Feb 3 at 2:39
jgonjgon
16.6k32244
16.6k32244
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
add a comment |
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
$begingroup$
Thank you @jgon
$endgroup$
– Ufomammut
Feb 3 at 2:47
add a comment |
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