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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.

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    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.2012-05-08

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