Is $E\ni x\mapsto(x,x)$ only measurable if $E$ is a Polish space?

Question

Let $(E,\mathcal E)$ be a measurable space and $$f:E\to E\times E\;,\;\;\;x\mapsto(x,x).$$ I've read that $f$ is $(\mathcal E,\mathcal E\otimes\mathcal E)$-measurable, if $E$ is a Polish space and $\mathcal E=\mathcal B(E)$. I don't get why $E$ needs to be a Polish space. If $\pi_i:E\times E\to E$ denotes the projection onto the $i$th coordinate, then $\pi_i\circ f=\text{id}_E$ is obviously $(\mathcal E,\mathcal E)$-measurable. So, $f$ should be $(\mathcal E,\mathcal E\otimes\mathcal E)$-measurable. What am I missing?