# Spotting the difference and improving rigor

What is the difference between these 2 questions?

I have been asked to prove the following 2 cases -- (1), (2):

2 maps $f_1,f_2$, where $f_1:X\to Y, f_2:Y\to X$ and $f_1f_2$ is the identity map,  are continuous. We are also given that $Y$ is (1) connected (2) path-connected and all pre-images of the elements in $Y$ are (1) connected (2) path-connected.

I hope to show that $X$ is (1) connected (2) path-connected.

I don't understand what the difference is in proving the 2 cases.

What I think is:

The preimage of a point $y\in Y$ wrt $f_1$ is simply the image of $f_2$. Since continuity preserves (path-)connectedness, points in $Y$ are (path-)connectedness implies that the images of such points under $f_2$ are  (path-)connected. This means that the pre-images of the points under $f_1$  can be joined by some path. Since we are given that the pre-images are  (path-)connected, it follows that the entire $X$ is  (path-)connected.

Please point out any flaws in the argument!

Also, it would be nice if someone would point out if the significance of the condition "all pre-images of the elements in $Y$ are (1) connected (2) path-connected." is as I guessed in the comments...

Thank you.

• I'm a little confused by the statement here - the fact that preimages of $Y$ under $f_1$ are (path-)connected follows from the fact that $f_1^{-1}(x)=\{f_2(x)\}$ as you point out in the first two sentences of the proof, so it doesn't seem like it's a necessary assumption. Secondly, what do you mean by two points being "connected"? I assume it means they lie in the same connected component (this is not the same as being connected by a path, which is where I imagine the ultimate confusion lies), but it's not terminology I'm familiar with. – mdp May 8 '12 at 19:18
• @MattPressland: Thabks for replying! 1) that has crossed my mind too, I guess that they want to ensure that $\{f_2(x)\}$ is (path-)connected in itself?? Not sure though... 2) still thinking about this one... – P. Willetts May 8 '12 at 19:31
• Well, that's always true, it's just a point because $f_1$ is injective. I would suggest trying to prove that if $Y$ is (path-)connected and if $f_1,f_2$ are maps as given, that $X$ is (path-)connected, and forget abut this second condition. – mdp May 8 '12 at 20:03