If $f:X_1 \rightarrow X_2$ and $g:X_2 \rightarrow X_3$ are homomorphisms. If $g \circ f =0$ does it imply that $Im f \subseteq ker g$? and how to show that? do you have an example? thanks :)
Asked
Active
Viewed 215 times
4
-
1Do you know the definitions of $\text{ker}$ and $\text{im}$? – Git Gud Apr 09 '13 at 20:05
-
@Danial Note that this is a necessary and sufficient condition so you can prove also that $Im f\subset Ker g$ implies $g\circ f$ is trivial. – Apr 09 '13 at 21:28
3 Answers
3
Take an element $y\in Im f$. Then $y$ is of the form $y=f(x)$, for some $x\in X_1$. Now apply $g$ to $y$. You get $g(y)=g(f(x))=(g\circ f)(x)$, which is zero by assumption. As an example, take $f$ arbitrary and $g(y)=0$ for every $y\in X_2$.
Federica Maggioni
- 8,422
-
-
-
-
2@Danial no, it's not, because $g$ sends everything to zero in any case. Conversely, if you assume that $f$ is onto, then $Im f=X_2=Ker(g)$, so that $g$ must be the constant morphism sending everything to zero – Federica Maggioni Apr 09 '13 at 20:19
-
1
-
@DavideGiraudo Of course, but usually I refrain from $+1$ing an answer which has been given before. Not in this case though. – Git Gud Apr 09 '13 at 20:27
2
Things in $Im(f)$ look like $f(x)$, any $y$ such that $g(y)=0$ is in the kernel of $g$, and $g(f(x))=0$. You have everything you need.
rschwieb
- 153,510
-
1@Danial $f$ is onto $Im(f)$ by definition. It is irrelevant if $f$ is not onto the entire $X_2$. – rschwieb Apr 09 '13 at 20:17