# Is the constant map a continuous function?

I've been set a question in an assignment which reads:

"Check whether the following functions are continuous or open. Check whether they are a homeomorphism.

$\dots$

$b)$ the constant map $f:X \rightarrow Y$ defined by $f(x)=y_0$ for some $y_0 \in Y$."

I've been given no more context than this. In other questions like this, I was told the topologies on the sets $X$ and $Y$. As far as I know, a function is a homeomorphism if both the function and its inverse are continuous and the map is bijective. A map function being continuous means that the preimage of any open set is open, so continuity of the inverse is the condition that the image of any open set is open.

So the fact that I've not been told the topologies on these sets $X$ and $Y$ makes me think that (whether or not this function is a homeomorphism) is independent of the topologies on the two sets. But I'm not sure how to show this. Thanks for any help!

A constant function is always continuous since the empty set and the whole space are its only candidates for preimages, and both are open.

A constant function is not necessarily open. It is if and only if in this context $\{y_0\}$ is an open set.

Homeomorphisms are bijective. A constant function can only be bijective if its domain and codomain are both singletons. In that case the function, and also its inverse are automatically continuous, so then indeed we deal with a homeomorphism. So indeed you don't need the topologies here. As said domain and codomain are singletons and on a one-point set there is only one topology.

• Ah ok so we need $\{y_0\}$ to be open. But there is a case where $X$ is a singleton and $\{y_0\}=Y$ is open hence the constant function is a homeomorphism? Edit: Having reread your post that is exactly what you said! Thanks a lot! Aug 17 '14 at 12:38

A function $f$ is continuous iff the inverse image by $f$ of any open set of $Y$ is an open of $X$. Use the fact that both $\emptyset$ and $X$ are open sets of $X$.

Also, unless $X$ is a singleton, your function is not bijective, a fortiori not a homeomorphism.

• For any $y\neq y_0$, $f^-1(\{y\})=\emptyset$, and $f^{-1}(\{y_0\})=X$ , these are both open. So I guess that for any $V$ open in $Y$, $f^{-1}(V)$ is either $X$ or $\emptyset$, which is open, so $f$ must be continuous. But my problem is that given any non-empty open set $U$ of $X$, $f(U)=\{y_0\}$ which is not necessarily open? Aug 17 '14 at 12:32
• Ok I see that it can't in general be invective though, so that rules out its being a homeomorphism. Thanks! Aug 17 '14 at 12:35