Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a problem from Dugundji's book, page $156$.

Let $f: X \rightarrow Y$, $g: Y \rightarrow X$ be continuous such that $g \circ f=1_{X}$. Prove that:

  1. If $Y$ is Hausdorff then so also is $X$ (done)

  2. $f(X)$ is closed in $Y$

I don't see how to show 2. Can you please help?

share|improve this question
3  
Can you write $f(X)$ as $g^{-1}(F)$ for some closed set $F$? –  GEdgar Jun 22 '11 at 0:10
    
@GEdgar: I suspect you are thinking on the diagonal right? not sure still how to involve it. –  user10 Jun 22 '11 at 1:15
add comment

2 Answers

up vote 5 down vote accepted

Here's an alternative approach more in keeping with Chap. VII of Dugundji. Let $h = f \circ g : Y \to Y$; $h$ is continuous, its range is $f[X]$, and $h \restriction f[X] = 1_{f[X]}$. Clearly $h(y) \neq y$ for $y \in Y \setminus f[X]$, so $f[X] = \{y \in Y:h(y) = 1_Y(y)\}$; $h$ and $1_Y$ are continuous, and $Y$ is Hausdorff, so you can now apply Dugundji's Thm. VII.1.5(1) to get the desired conclusion.

share|improve this answer
    
thanks! –  user10 Jun 23 '11 at 3:33
add comment

Let $y_\alpha$ be a net in $f(X)$ with limit $y \in Y$. Then $y_\alpha = f(x_\alpha)$ for a unique $x_\alpha \in X$ (since $f$ is one-to-one, by the assumption) and for each $\alpha$. Then by continuity of $g$, we have $g(y_\alpha) \to g(y)$ but also $g(y_\alpha)=g(f(x_\alpha))=x_\alpha$, so $x_\alpha \to g(y)$. Now by continuity of $f$ we have $y_\alpha = f(x_\alpha) \to f(g(y))$.If $Y$ is Hausdorff, then we have uniqueness of limits and therefore $f(g(y))=y$, so in particular $y \in f(X)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.