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...