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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?

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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
up vote 6 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.

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thanks! – user10 Jun 23 '11 at 3:33

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 $y \in f(X)$.

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