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Here is the definition of Markov kernel I am reading from this paper, https://arxiv.org/pdf/1410.5110.pdf, it defines the Markov kernel as:

A Markov kernel, $\tau$, is a map from an element of the sample space and the $\sigma$-algebra to a probability:

$\tau: Q \times \mathcal{B}(Q) \rightarrow [0,1]$

such that the kernel is a measurable function in the first argument,

$\tau(\cdot, A):Q\rightarrow Q, \forall A \in \mathcal{B}(Q)$,

and a probability measure in the second argument,

$\tau(q, \cdot): \mathcal{B}(Q) \rightarrow [0, 1], \forall q \in Q$.

My question is about the first claim, why it is a map from $Q \rightarrow Q$?

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    Actually, $\tau(\cdot, A):Q\rightarrow [0,1]$, for all $A \in \mathcal{B}(Q)$. – 2017-02-19
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    @Did Thanks. Yes, I think it should be $\tau(\cdot, A): Q \rightarrow [0,1]$, my understanding is that, give an event $A$, then the map is still a probability measure that from sample space $Q$ to $[0, 1]$. – 2017-02-19
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    No, the map $\tau(\cdot,A)$ is not a probability measure on $Q$, the most one can say is that every $\tau(q,A)$ is in $[0,1]$. – 2017-02-19
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    My understanding is that, when $A$ is constant, then $\tau(\cdot, A ): Q \rightarrow [0, 1]$, then why we can not say $\tau(\cdot, A )$ is a probability measure on $Q$? Assuming we have a discrete markov chain with transition matrix, let's say state $X_n = 1$, for example, then we can get the probability of $X_{n-1}$ from the transition matrix, is this understanding correct? – 2017-02-19
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    To begin with, $\tau(\ ,A)$ is not defined on a sigma-algebra on $Q$, but on $Q$ itself. // What would be a source for the claim in your post? – 2017-02-19
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    https://arxiv.org/pdf/1410.5110.pdf, page 3. Thanks so much. – 2017-02-19
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    A typo, **obviously**. – 2017-02-19

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Yeah, I think there is a typo over here.

It should be $Q \to [0, 1]$

And, it's quite reasonable since for any given set A, $\tau(\cdot,A)$ is a map from $Q$ to $[0, 1]$