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Let $HAUSS$ be the category of all Hausdorff spaces and $i:\mathbb{Q}\rightarrow \mathbb{R}$ be the inclusion map. I am studying a book on Homology theory. It is mentioned there, that the map $i$ is monomorphism as well as epimorphism because $\mathbb{Q}$ is dense in $\mathbb{R}$. But it is not an isomorphism.

My question is that, it can be clearly seen that $i$ is not an homeomorphism (hence isomorphism) but how can we say that $i$ is an epimorphism using the densness of $\mathbb{Q}$. Where in category theory a map $f$ will be an epimorphism if it can be right cancelled.

MANI
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3 Answers3

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More generally, in the category of Hausdorff spaces, a map $f\colon X\to Y$ is an epimorphisms if and only if $f(X)$ is dense in $Y$.

In fact, this property characterizes the Hausdorff spaces; namely, $Y$ is Hausdorff if and only if for every space $X$ and every continuous morphism $f\colon X\to Y$, $f$ is completely determined by its value on a dense subset of $X$. See this question for the “only if” clause, and this one (and references therein) for the converse.

Arturo Magidin
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Suppose $f\circ i = g\circ i$, $f,g: \mathbb R\to X$ where $X$ is Hausdorff.

Then $\{x \in \mathbb R\mid f(x) = g(x)\} \supset \mathbb Q$ by hypothesis, and it's closed in $\mathbb R$ (because it's $h^{-1}(\Delta_X)$, where $h: \mathbb R\to X\times X, x\mapsto (f(x),g(x))$ and $\Delta_X = \{(y,y)\in X\times X\}$ , and $\Delta_X$ is closed in $X\times X$ by Hausdorffness), therefore it contains $\overline{ \mathbb Q} = \mathbb R$.

So $f(x) = g(x)$ for all $x\in \mathbb R$, in other words $f=g$.

Maxime Ramzi
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  • thanks for your answer. Can you explain a little in categorical frame that why $i$ is not an isomorphism? – MANI Sep 19 '19 at 06:34
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    @MANISHANKARPANDEY: Because it is not bijective; isomorphisms in concrete categories must be bijective, since they have inverses. – Arturo Magidin Sep 19 '19 at 06:35
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This is basically saying that a continuous map $f:\Bbb R\to X$ for a Hausdorff space is determined by its values on $\Bbb Q$.

To see why this is true, if $a\in \Bbb R$ the're a sequence $(b_n)$ in $\Bbb Q$ with limit $a$. By continuity $f(a)=\lim_{n\to\infty }f(b_n)$ (noting that limits in Hausdorff spaces are unique). So $f(a)$ is determined by the $f(b)$ with $b\in\Bbb Q$.

Therefore if $f$, $g:\Bbb R\to X$ are continuous and $f|_{\Bbb Q}=g|_{\Bbb Q}$ then $f=g$. This means that $i:\Bbb Q\to\Bbb R$ is an epimorphism in the category of Hausdorff spaces.

Angina Seng
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