If $\varphi : X\times Y\rightarrow Z$ is continuous, $\widehat{\varphi} : X\rightarrow C(Y,Z)$ is not necessary continuous.

Question

Let $X,Y,Z$ be metric spaces. If we have continuous map $\varphi : X\times Y \rightarrow Z$, we can get continuous map for each $x\in X$ $\widehat{\varphi}(x) : Y\rightarrow Z$ defined by $\widehat{\varphi}(x)(y)=\varphi(x,y)$. It is proven in If the function $\varphi \colon Z\rightarrow C(X,Y)$ is continuous then $F\colon Z\times X\rightarrow Y$, $F(z,x)=\varphi (z)(x)$ will be continuous. if $Y$ is locally compact, we even get $\widehat{\varphi} : X \rightarrow C(Y,Z)$ is continuous where $C(Y,Z)$ is set of all continuous functions from $Y$ to $Z$. My question is, if $Y$ is not locally compact, can $\widehat{\varphi}$ is also continuous? My lecturer said there is counterexample but he did not mention one.

Answer 1

Note that all spaces below are Hausdorff (not sure if this is needed, but I'm assuming this to not state false theorems).

We consider $C(Y, Z)$ with compact-open topology here.

There is a misunderstanding, if $\varphi:X\times Y\to Z$ is continuous, then $\hat\varphi:X\to C(Y, Z)$ is always continuous. It's the implication $\hat\varphi$ continuous $\implies$ $\varphi$ continuous that isn't always true.

Namely, witness the following proposition (see e.g. Dugundji's Topology):

Proposition. If $\hat\varphi:X\to C(Y, Z)$ is continuous and $Y$ is locally compact (or $X\times Y$ is a $k$-space), then $\varphi:X\times Y\to Z$ is continuous.

Since every metric space is a $k$-space, in your setting of metric spaces we have:

Theorem. If $X, Y, Z$ are metric spaces then $\varphi$ is continuous iff $\hat\varphi$ is continuous.

So actually both implications are always true!

Edit: Since the OP was asking for it, here's a counter-example when we're not in the real of metric spaces. Let $Y$ be a Tychonoff space that's not locally compact, for example $Y = \mathbb{Q}$, say $y_0\in Y$ doesn't have a compact neighbourhood. If we take $\varphi:C(Y, [0, 1])\times Y\to [0, 1]$ to be the evaluation map, that is $\varphi(f, y) = f(y)$ then $\varphi$ is not continuous. The proof is relatively easy:

Proof: Suppose otherwise, that $\varphi$ is continuous, let $0:Y\to [0, 1]$ denote tha map $0(x) = 0$, then there are neighbourhoods $U$ of $y_0$ and $V = \bigcap_{i=1}^n (K_i, V_i)$ of $0$ such that $\varphi[U\times V]\subseteq [0, 1)$ (from continuity of $\varphi$ at $(0, y_0)$). Here $K_i\subseteq Y$ are compact, and $V_i\subseteq [0, 1]$ are open.

Since $\overline{U}$ is not compact, we can't have $U\subseteq \bigcup_{i=1}^n K_i$, so that there is $y_1\in U\setminus \bigcup_{i=1}^n K_i$. Since $Y$ is Tychonoff we can find a continuous function $h:Y\to [0, 1]$ such that $h[\bigcup_{i=1}^n K_i] = \{0\}$ and $h(y_1) = 1$. Then $(h, y_1)\in V\times U$ but $\varphi(h, y_1) = 1$, which is impossible. The contradiction shows that $\varphi$ can't be continuous at the point $(0, y_0)$. $\square$

Here $Z = [0, 1]$ and $X = C(Y, [0, 1])$. The corrsponding map $\hat\varphi:C(Y, [0, 1])\to C(Y, [0, 1])$ is just the identity map $\text{Id}_{C(Y, [0, 1])}$ and as such is continuous.

This shows that from continuity of $\hat\varphi$ it doesn't follow that $\varphi$ is continuous.