How to show $(gf)^{-1} = f^{-1}g^{-1}$? Suppose that $f:A\rightarrow B$ and $g:B \rightarrow C$ are both one-to-one and onto. Prove that $gf$ is one-to-one and onto. Prove further that $(gf)^{-1} = f^{-1}g^{-1}$.   I have already proven the first part, but the second part has always puzzled me. I have tried assuming $x \in (gf)^{-1}$ but that doesn't lead to nowhere. Nor does $x \in (gf)^{-1}(t)$ and showing $x = t$. How to prove?
How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$? (inverse of composition) I'm doing exercise on discrete mathematics and I'm stuck with question:     If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$.   I've no idea how to prove this, please help me by give me some reference or hint to its solution.
Prove if f(z) is entire function s.t $|f(z)|\leq M|z|$ then $f(z)$ Is polynomial of degree $1$ Let $f(z)$ be an entire function s.t $|f(z)|\leq M|z|$ ; $M&gt;0$ is a constant then $f(z)$ Is polynomial of degree 1   My try $f(z)$ is entire so it can be written as a $\sum a_n z^n$  $$|\sum a_n z^n|\leq \sum |a_n||z|^n\leq M|z|$$  So $n=1$  Seems that the move $\sum |a_n||z|^n\leq M|z|$ is problematic How to apply a macro to each column of a table Is there a nice example of integral domain $R$ and $s,t \in R$, with $Rs \cap Rt$ not principal but lcm($s,t$) exists? I am currently searching for an integral domain $R$ which allows lcm, but where the intersection of two principal ideals is not necessarily a principal ideal (or at least an example with two certain elements).   As converse example (where the sum of two ideals is not principal but $R$ allows gcd) is $\mathbb Z[X]$. It allows gcd, but obviously $\left&lt;X+1,2\right&gt;$ is not principal.  An example of an integral domain where the intersection of two principal ideals is not principal would be $\mathbb Z[\sqrt{-5}]$ with the ideal $\left&lt;2\right&gt; \cap \left&lt;1+\sqrt{-5}\right&gt;$, but this doesn't allow lcm.  Since I can't find a nice example, I put the question here. Maybe someone has already seen this? (I am not completely sure, one even exists, but pretty certain there is an example. In the case my hunch is wrong, I would be happy, if someone could proof me wrong.) How to resolve the "GPT detected. Please create a BIOS-Boot partition" error when using boot-repair? After installing Ubuntu, Windows didn't boot so I installed boot repair to fix this problem. But I am getting the message displayed below.  I tried to make a new partition with unformatted file system but still receiving the same message.  How do I fix it? Prove that for every $a, b\in R\setminus\{0\}$ is correct this inequality:  $$\frac{|a+b|}{1+|a+b|}&lt;\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$ Matrix of Linear Transformation by right multiplication I am trying to solve the following problem:   Let $A$ be an $n\times n$ matrix, and let $V$ denote the space of $n$-dimensional row vectors. What is the matrix of the linear operator ‘‘right multiplication by $A$’’ with respect to the standard basis of $V$?  I am not sure where to begin with this problem. Zero dimensional points do not take up space, so then wouldn't everything in the universe be literally empty? Or is there something that I'm missing? I have a VIP skybag (I believe this does not matter). It has a 3-number TSA lock built into the suitcase. I know the combination to unlock, let's say it's 345. Recently it was locked and while there was a different number on the combination lock the reset/set lock button was pressed.  Since then the reset/set lock button is pressed (stuck) and the lock button also does not unlock. Any pointers on how to unlock it? I am thinking of unscrewing the lock from inside of the suitcase but it seems quite a hassle, as I may have to assemble it back. Let me know if there is any other option to unlock it from the outside. cron job schedule Can a coin with an unknown bias be treated as fair? No connection could be made because the target machine actively refused it (PHP / WAMP) Is there a way to export each rendered layer as its own image? My scene uses multiple render layers.  I'd like to edit each rendered layer in Photoshop, and composite it in Photoshop.  Is there a way to export each layer as an individual image? Could authors overturn recent community review decisions? When new users edit a post, the review is posted to a review queue and other reputable users can review the edit and choose to approve or reject it. However, I feel that a lot of people do it pretty quickly.    to . He was fixing broken links in the answer, and as soon as I saw it I knew it would have been a good edit. However, only one reviewer found so, and three others rejected it. I ended up manually doing the edit, and poor Onato didn't get any credit for it (in fact, he got an undeserved 'rejected' for it).  The point I want to bring is that, first and foremost, the person who wrote the answer probably knows best if an edit is acceptable or not. Could there be a time frame where the author of the answer (or question) can overturn a review decision?    EDIT: one more  was rejected by overeager reviewers as well. I absolutely disagree with the notion that the community knows better. The community repeatedly proves that it only wants Internet points.    EDIT: I told a commenter to fix something himself in one of my posts, but . I went back and applied the changes myself. It's not credited to him and his stats got a rejected edit now (which really should be an approved edit). The reason ("This edit is incorrect or an attempt to reply to or comment on the existing post") is blatantly wrong and shows once more that the community has much less insight into edit quality than the people who wrote the answers in the first place. It's true that the UI didn't make it easy for them, but those guys clearly didn't read the comment thread; I did, because as the author, I was notified.    EDIT: This time around, a user fixed a blatant (though small) mistake in one of my accepted answers. One hour later, another, unrelated 10k user , even though it had been approved by the community. I went back to undo his undo.  This goes on again to show that an active user is the best judge for the quality of an edit, and that since I can have the last word on edits on my own posts, I should be allowed to retroactively overturn a community edit vote and accordingly credit the person who proposed the change.    EDIT I wrote a  years ago. Some guy found it and tried to fix it but the community thought that it wasn't worth it. I think that the changes are great and once more I'll integrate them by myself into my post. Unfortunately for this user, he won't be credited for it. I still don't understand why I can't have a binding vote on edits on my posts. In Python, I can compile a regular expression to be case-insensitive using re.compile:  &gt;&gt;&gt; s = 'TeSt' &gt;&gt;&gt; casesensitive = re.compile('test') &gt;&gt;&gt; ignorecase = re.compile('test', re.IGNORECASE) &gt;&gt;&gt;  &gt;&gt;&gt; print casesensitive.match(s) None &gt;&gt;&gt; print ignorecase.match(s) &lt;_sre.SRE_Match object at 0x02F0B608&gt;   Is there a way to do the same, but without using re.compile. I can't find anything like Perl's i suffix (e.g. m/test/i) in the documentation. What is the name of the problem where lines appear in a photo where there should just be a smooth gradient? How to show full date and time in Nautilus/Files 3.6+ list view? laptop fails to suspend or hibernate (wakes up after 2-5 seconds without user interaction) How to fix Eclipse with OpenJDK 9 How do I use the Chinese remainder theorem to find all the square roots of 11 in $\mathbb Z_{35}$? If  $\mathbb R^*$ and $\mathbb C^*$ be respectively multiplicative groups of non-zero real and complex number then whether $\operatorname{Hom}(\mathbb Z_6,R^*\oplus C^*)$  is  isomorphic to $\mathbb Z_{12}$? or $\mathbb Z_6$? or $\mathbb Z_6\oplus \mathbb Z_2$?