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Let $G,H,K$ finite group. Why if we define a morphism group $f:G/H\to K$, we have to show that $xH=yH\implies f(xH)=f(yH)$ ? btw how can we have $xH=yH$ and $f(xH)\neq f(yH)$, it looks very weird...

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Usually, the easiest way to give a definition of a group homomorphism $f:G/H \to K$ is by defining what happens to the coset $xH$ where we choose a coset representative $x$. For example, we often define $f(xH) = g(x)$ for some other group homomorphism $g:G \to K$. But remember, $G/H$ is a space of cosets, so $x$ is not a unique representative of the coset $xH$. So it's possible that $xH = yH$ but $g(x) \neq g(y)$, in which case $f$ is not a function, since it doesn't output a unique value for the coset $xH$.

Here's a concrete example: Let $G = \mathbb{Z}$ and $H = 2\mathbb{Z}$ and $K = \mathbb{Z}$. Say we try to define $f:G/H \to K$ by $f(x+H) = x$. We check that $f$ is a homomorphism by the calculation \begin{equation*} f((x+H) + (y+H)) = f((x+y)+H) = x+y = f(x+H) + f(y+H) \end{equation*} so $f$ is a group homomorphism, right? Wrong. We haven't checked that $f$ is well-defined, that is, we haven't check that $x+H = y+H \implies f(x+H) = f(y+H)$. And in fact, this is false. We have $2+H = 4+H$, but $f(2+H) \neq f(4+h)$ since $2 \neq 4$.

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    I like this answer, but I think that first sentence needs some tweaking after "$f: G/H \to K$"; I'm not sure you explicitly say that we define $f(xH)$ by defining what happens to *a representative* (or element) of $xH$, although it is implied. – 2017-01-29
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    @pjs36 Thanks, I added clarification. – 2017-01-29
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If $G$ and $K$ are groups and $H$ is a normal subgroup of $G$, then let $f:G/H\to K$. Since $G/H$ is actually a set of equivalence classes $xH$ such that $xH = yH$ if and only if $x^{-1}y\in H$, we need to ensure that the evaluation $f(xH)$ is equal to the evaluation $f(yH)$ whenever $x$ and $y$ are representatives of the same equivalence class. This is just the statement that the function $f$ is well-defined.